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20 Quadratic Equations

Topics Covered

Brief Overview of Polynomial Equations

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

Polynomial Equation

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We have already solved polynomial equations of degree one. Polynomial equations of degree one are linear equations and of the form ax + b = c.

We are now going to solve polynomial equations of degree two. A polynomial equation of degree two is called a quadratic equation. Listed below are some examples of quadratic equations:

Ex. 1 Ex. 2 Ex. 3 Ex. 4
x2 + 5x + 6 = 0 3y2 + 4y = 10 64u2 − 81 = 0 n(n + 1) = 42

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get n2 + n.

Quadratic Equation

An equation of the form ax2 + bx + c = 0 is called a quadratic equation.

a, b, and c are real numbers and a≠0

(If a = 0, then 0 · x2 = 0 and we are left with no quadratic term.)

Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax2. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

Zero Product Property

We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Zero Product Property

If a · b = 0, then either a = 0 or b = 0 or both.

We will now use the Zero Product Property, to solve a quadratic equation.

How to use the zero product property.
  1. Set each factor equal to zero.
  2. Solve the linear equations.
  3. Check.

Try it! – How to Solve a Quadratic Equation Using the Zero Product Property

Solve: (5n − 2)(6n − 1) = 0.

Solution


The equation is open parentheses 5n minus 2 close parentheses open parentheses 6n minus 1 close parentheses equals 0. The product equals zero, so at least one factor must equal zero. Step 1 is set each factor equal to zero. So, 5n minus 2 equals 0 and 6n minus 1 equals 0.Step 2 is to solve the linear equations. So, we get n equal to 2 by 5 and n equal to 1 by 6.Step 3 is to check by substituting each solution separately into the original equation.

Solve Quadratic Equations by Factoring

The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we start with the quadratic equation in standard form, ax2 + bx + c = 0. Then we factor the expression on the left.

How to solve a quadratic equation by factoring.
  1. Write the quadratic equation in standard form, ax2 + bx + c = 0.
  2. Factor the quadratic expression.
  3. Use the Zero Product Property.
  4. Solve the linear equations.
  5. Check. Substitute each solution separately into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form.

Try it! – How to Solve a Quadratic Equation by Factoring

Solve: 2y2 = 13y + 45.

Solution


The equation is 2 y squared equals 13y plus 45. Step 1 is to write it in standard form a x squared plus bx plus c. So we have 2 y squared minus 13y minus 45 equals 0.Step 2 is to factor the quadratic expression. So we have 2y plus 5, y minus 9 equals 0.Step 3 is to use the zero product property. Setting each factor equal to zero, we have two linear equations: 2y plus 5 equals 0 and y minus 9 equals 0.Step 4 is to solve the linear equations. We get, y equals minus 5 by 2 and y equals 9.Step 5 is to check by substituting each solution separately into the original equation Substitute y equals -five over two and yequals 9 into the original equation. Start with two times -5/2 squared equal 13 ( times -5/2) + 45. Does this equal each other? Yes, because twenty-five over 2 equlas twenty-five over two. Substitue y equals 9 in. You start with 2 times (9) squared equals 13 times 9 + 45. Does this equal each other? Yes, because 162 equals 162.

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in earlier sections! Do you recognize the special product pattern in the next example?

Try it!

Solve: 169q2 = 49.

Solution

Steps Algebraic
Write the quadratic equation in standard form. 169q2 – 49 = 0 NA
Factor. It is a difference of squares. (13q – 7) (13q + 7) = 0 NA
Use the Zero Product Property to set each factor to 0. 13q – 7 = 0
 13q + 7 = 0
Solve each equation. 13q = 7 13q = -7
Solutions q = \displaystyle\frac{7}{13} q = –\displaystyle\frac{7}{13}

Check: We leave the check up to you.
 In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Try it!

Solve: (3x − 8)(x − 1) = 3x.

Solution

Steps Algebraic
Equation (3x − 8)(x − 1) = 3x
Multiply the binomials. 3x2 − 11x + 8 = 3x
Write the quadratic equation in standard form. 3x2 − 14x + 8 = 0
Factor the trinomial. (3x − 2)(x − 4) = 0
Use the Zero Product Property to set each factor to 0 3x − 2 = 0      x − 4 = 0
Solve each equation 3x = 2 x = 4
\displaystyle x=\frac{2}{3}

 Check your answers.

 The check is left to you. 

In the next example, when we factor the quadratic equation, we will get three factors. However, the first factor is a constant. We know that factor cannot equal 0.

Try it!

Solve: 3x2 = 12x + 63.

Solution

Steps Algebraic
Equation 3x2 = 12x + 63
Write the quadratic equation in standard form. 3x2 − 12x − 63 = 0
Factor the greatest common factor first. 3(x2 − 4x − 21) = 0
Factor the trinomial. 3(x − 7)(x + 3) = 0
Use the Zero Product Property to set each factor to 0. 3 ≠ 0    x − 7 = 0    x + 3 = 0
Solve each equation. 3 ≠ 0    x = 7    x = −3
Check your answers. The check is left to you.

Solve Quadratic Equations of the form ax2 = k using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x2 = 9.

Steps Algebraic
Equation x2 = 9
Put the equation in standard form. x2 − 9 = 0
Factor the difference of squares. (x − 3)(x + 3) = 0
Use the Zero Product Property. x − 3 = 0  and x + 3 = 0
Solve each equation. x = 3    and x = −3

We can easily use factoring to find the solutions of similar equations, like x2 = 16 and x2 = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, x = 4, x = −4 and x = 5, x = −5.

But what happens when we have an equation like x2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13)2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

If n2 = m, the n is a square root of m.

Since these equations are all of the form x2 = k, the square root definition tells us the solutions are the two square roots of k. This leads to the Square Root Property.

Square Root Property

If x2 = k, then

x = \sqrt{k} or x = −\sqrt{k} or x = ±\sqrt{k}.

Notice that the Square Root Property gives two solutions to an equation of the form x2 = k, the principal square root of k and its opposite. We could also write the solution as x = ±\sqrt{k}. We read this as x equals positive or negative the square root of k.

Now we will solve the equation x2 = 9 again, this time using the Square Root Property.

Algebraic
x2 = 9
x = ±\sqrt{9}
x = ±3
So x = 3 or  x = −3.

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x2 = 7.

Algebraic
x2 = 7
x = \sqrt{7}, x = −\sqrt{7}

We cannot simplify [latex]\sqrt{7}[/latex], so we leave the answer as a radical.

How to solve a quadratic equation using the square root property.
  1. Isolate the quadratic term and make its coefficient one.
  2. Use Square Root Property.
  3. Simplify the radical.
  4. Check the solutions.

Try it!

How to solve a Quadratic Equation of the form ax2 = k Using the Square Root Property

Solve: x2 − 50 = 0.

Solution
Step one is to isolate the quadratic term and make its coefficient one. For the equation x squared minus fifty equals zero, first add fifty to both sides to get x squared by itself. The new equation is x squared equals fifty.Step two is to use Square Root Property. Remember to write the plus or minus symbol. The equation created is x equals the positive or negative square root of 50.Step three is to simplify the radical if possible. Continue to write equivalent equations. X equals the positive or negative square root of twenty five times the square root of five. X equals positive or negative five times the square root of five. Rewrite to show two solutions: x equals five times the square root of five or x equals negative five times the square root of five.Step four is to check the solutions. Substitute x equals five times the square root of five and or x equals negative five times the square root of five into the original equation. Start with x equals five times the square root of five. x squared minus fifty equals zero. Does the square of five times the square root of five minus fifty equal zero? Does twenty five times two minus fifty equal zero? Yes, because zero equals zero. Next check x equals negative five times the square root of five. x squared minus fifty equals zero. Does the square of negative five times the square root of five minus fifty equal zero? Does twenty five times two minus fifty equal zero? Yes, because zero equals zero.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Try it!

Solve: 3z2 = 108.

Solution

Steps Algebraic
Equation 3z2 = 108
The quadratic term is isolated.

Divide by 3 to make its coefficient 1.		 \displaystyle\frac{3z^2}{3} = \displaystyle\frac{108}{3}
Simplify. z2 = 36
Use the Square Root Property. z = ± \sqrt{36}
Simplify the radical. z = ±6
Rewrite to show two solutions. z = 6, z = −6
Check the solutions:

 Check the solutions by substituting each value into the original equation, three times z squared equals one hundred eight. Start with z equals six. Substituting yields three times the square of six, or three times thirty six. Three times thirty six equals one hundred eight, so z equals six is a solution of the original equation. Next substitute z equals negative six. Three times the square of negative six is equivalent to three times thirty six, or one hundred eight. z equals negative six is also a solution.

The Square Root Property states ‘If x2 = k,’ What will happen if k < 0? This will be the case in the next example.

Try it!

Solve: x2 + 72 = 0

Solution

Steps Algebraic
Equation x2 + 72 = 0
Isolate the quadratic term. x2 = −72
Use the Square Root Property. x = ±\sqrt{-72}
Simplify using complex numbers. x = ±\sqrt{72}i
Simplify the radical. x = ±6\sqrt{2}i
Rewrite to show two solutions. x = 6\sqrt{2}i, x = −6\sqrt{2}i
Check the solutions:

 Start with the equation three times z squared equals one hundred eight. The quadratic term is isolated, so divide by 3 to make its coefficient 1. Three times z squared divided by three equals one hundred eight divided by three. Simplify to find that z squared equals thirty six. Use the Square Root Property to see that z equals positive or negative square root thirty six. Simplify the radical. Z equals positive or negative six. Rewrite to show two solutions. z equals six or z equals negative six. Check the solutions by substituting each value into the original equation, three times z squared equals one hundred eight. Start with z equals six. Substituting yields three times the square of six, or three times thirty six. Three times thirty six equals one hundred eight, so z equals six is a solution of the original equation. Next substitute z equals negative six. Three times the square of negative six is equivalent to three times thirty six, or one hundred eight. z equals negative six is also a solution.

Our method also works when fractions occur in the equation, we can solve any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

Try it!

Solve: \displaystyle\frac{2}{3}u2 + 5 = 17.

Solution

Steps Algebraic
Equation \displaystyle\frac{2}{3}u2 + 5 = 17
Isolate the quadratic term. Isolate the quadratic term to yield two thirds u squared equals twelve.
Multiply by \displaystyle\frac{3}{2} to make the coefficient 1. The coefficient of u squared is two thirds, so multiply both sides of the equation by three halves. REsulting in 3/2 times (2/3 u squared) equals three halves times 12.
Simplify. Simplifying yields u squared equals eighteen. Use the Square Root Property to see that u equals the positive or negative square root of eighteen. Simplify the radical rewriting the square root of eighteen as the product of the square root of nine and the square root of two. Simplifying shows u equals positive or negative three times the square root of two. Rewrite to show two solutions: u equals three square root two or u equals negative three square root two. Check the solutions by substituting each value into the original equation, two thirds u squared plus five equals seventeen. Start with u equals three square root two. Substituting on the left side of the equation yields the sum of two thirds times the square of three square root two and five. We need to show that this expression equals seventeen. Simplifying the square gives the sum of two thirds times eighteen and five. Finding the product gives twelve times five, which is equal to seventeen. This value, three square root two, is a solution of the original equation. Next check the value u equals negative three square root two. Substitute this value for u in the original equation. Simplifying the square yields two thirds times eighteen plus five. We need to show that this expression equals seventeen. Two thirds times eighteen plus five equals twelve plus five, or seventeen. So negative three square root two a solution.
Use the Square Root Property. Use the Square Root Property to see that u equals the positive or negative square root of eighteen.
Simplify the radical. Simplify the radical rewriting the square root of eighteen as the product of the square root of nine and the square root of two.
Simplify. Simplifying shows u equals positive or negative three times the square root of two.
Rewrite to show two solutions. Rewrite to show two solutions: u equals three square root two or u equals negative three square root two.
Check:
 Check the solutions by substituting each value into the original equation, two thirds u squared plus five equals seventeen. Start with u equals three square root two. Substituting on the left side of the equation yields the sum of two thirds times the square of three square root two and five. We need to show that this expression equals seventeen. Simplifying the square gives the sum of two thirds times eighteen and five. Finding the product gives twelve times five, which is equal to seventeen. This value, three square root two, is a solution of the original equation. Next check the value u equals negative three square root two. Substitute this value for u in the original equation. Simplifying the square yields two thirds times eighteen plus five. We need to show that this expression equals seventeen. Two thirds times eighteen plus five equals twelve plus five, or seventeen. So negative three square root two a solution.

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.

Try it!

Solve: 2x2 − 8 = 41.

Solution

Steps Algebraic
Equation 2x2 − 8 = 41
Isolate the quadratic term. . Isolate the quadratic term to yield two times x squared equals forty nine.
Divide by 2 to make the coefficient 1. Divide both sides of the equation by two to make the leading coefficient one. Resulting in two x squared divided by 2, equals 49 divided by 2.
Simplify. Simplify so that the equation becomes x squared equals forty nine halves.
Use the Square Root Property. Use the Square Root Property to see that x equals the positive or negative square root of forty nine halves.
Rewrite the radical as a fraction of square roots. Simplify the radical rewriting the square root of forty nine divided by two as the quotient of the square root of forty nine and the square root of two.
Rationalize the denominator. Rationalize the denominator by multiplying both the numerator and denominator on the right side of the equation by the square root of two.
Simplify. Simplifying shows x equals positive or negative seven times the square root of two divided by two.
Rewrite to show two solutions. Rewrite to show two solutions: x equals seven square root two divided by two or x equals negative seven square root two divided by two. We leave the check for you.
Check: We leave the check for you.

Solve Quadratic Equations of the form a(x – h)2 = k using the Square Root Property

We can use the Square Root Property to solve an equation of the form a(x − h)2 = k as well. Notice that the quadratic term, x, in the original form ax2 = k is replaced with (x − h).

On the left is the equation a times x square equals k. Replacing x in this equation with the expression x minus h changes the equation. It is now a times the square of x minus h equals k.The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a, then the Square Root Property can be used on (x − h)2.

Try it!

Solve: 4(y − 7)2 = 48

Solution

Steps Algebraic
Equation 4(y − 7)2 = 48
Divide both sides by the coefficient 4. (y − 7)2 = 12
Use the Square Root Property on the binomial y − 7 = ±\sqrt{12}
Simplify the radical. y  − 7 = ±2\sqrt{3}
Solve for y. y = 7 ± 2\sqrt{3}
Rewrite to show two solutions. y = 7 + 2\sqrt{3}, y = 7 – 2\sqrt{3}
Check:
 Check the values in the original equation, four times the square of y minus seven equals forty eight. Start with y equals seven plus two times square root three. Substituting on the left side of the equation yields four times the square of the expression seven plus two square root three minus seven. We need to show that this expression equals forty eight. Simplifying the expression yields four times the square of two square root three. Simplifying the square gives four times twelve, which is equal to forty eight. This value, seven plus two times square root three, is a solution of the original equation. Next check the value y equals seven minus two times square root three. Substitute this value for y in the original equation. Four times the square of the expression seven minus two square root three minus seven simplifies to four times the square of negative two square root three. Simplify the square to yield four times twelve, or forty eight. So seven minus two times square root three.

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Try it!

Solve: \displaystyle\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}.

Solution

Steps Algebraic
Equation \displaystyle\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}
Use the Square Root Property. x − \displaystyle\frac{1}{3} = ± \sqrt{\displaystyle\frac{5}{9}}
Rewrite the radical as a fraction of square roots. x − \displaystyle\frac{1}{3} = ± \displaystyle\frac{\sqrt{5}}{\sqrt{9}}
Simplify the radical. x − \displaystyle\frac{1}{3} = ± \displaystyle\frac{\sqrt{5}}{3}
Solve for x. x = \displaystyle\frac{1}{3} ± \displaystyle\frac{\sqrt{5}}{3}
Rewrite to show two solutions. x = \displaystyle\frac{1}{3} + \displaystyle\frac{\sqrt{5}}{3},    x = \displaystyle\frac{1}{3} – \displaystyle\frac{\sqrt{5}}{3}
Check: We leave the check for you.

Sometimes the solutions are complex numbers.

Try it!

Solve: (2x − 3)2 = −12.

Solution

Steps Algebraic
Equation (2x − 3)2 = −12
Use the Square Root Property. 2x – 3 = ± \sqrt{-12}
Simplify the radical. 2x – 3 = ± 2 \sqrt{3i}
Add 3 to both sides. 2x = 3 ± 2 \sqrt{3i}
Divide both sides by 2. x = \displaystyle\frac{3 \pm 2 \sqrt{3i}}{2}  
Rewrite in standard form. x = \displaystyle\frac{3}{2} ±  \displaystyle\frac{2\sqrt{3i}}{2}
Simplify. x = \displaystyle\frac{3}{2} ±  \sqrt{3i}
Rewrite to show two solutions. x = \displaystyle\frac{3}{2} +  \sqrt{3i} ,  x = \displaystyle\frac{3}{2} –  \sqrt{3i}
Check: We leave the check for you.

The left sides of the equations in the next two examples do not seem to be of the form a(x − h)2. But they are perfect square trinomials, so we will factor to put them in the form we need.

Try it!

Solve: 4n2 + 4n + 1 = 16.

Solution

We notice that the left side of the equation is a perfect square trinomial. We will factor it first.

Steps Algebraic
Equation 4n2 + 4n + 1 = 16
Factor the perfect square trinomial. (2n + 1)2 = 16
Use the Square Root Property. 2n + 1 = ± \sqrt{16}
Simplify the radical. 2n + 1 = ±4
Solve for n. 2n = −1 ± 4
Divide each side by 2.  \displaystyle\frac{2n}{2} = \displaystyle\frac{-1\pm4}{2}
n = \displaystyle\frac{-1 \pm 4}{2} 
Rewrite to show two solutions. n = \displaystyle\frac{-1+4}{2}, n = \displaystyle\frac{-1 - 4}{2}
Simplify each equation. n =  \displaystyle\frac{3}{2},  n = –\displaystyle\frac{5}{2}
Check:

 Substitute each value into the original equation 4 times n squared plus 4 n plus 1 equals 16, to check. Start by substituting three halves for n on the left side of the equation. We want to show that this expression is equal to 16. 4 times the square of three halves plus 4 times three halves plus 1 equals 4 times nine fourths plus 4 times three halves plus 1. This expression simplifies to 9 plus 6 plus 1, or 16. Next substitute negative five halves into the original equation. 4 times the square of negative five halves plus 4 times five halves plus 1 equals 4 times 25 fourths plus 4 times negative five halves plus 1. This expression simplifies to 25 minus 10 plus 1, or 16. Negative five halves is a solution.

Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

 

Solve Quadratic Equations using Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.

We start with the standard form of a quadratic equation and solve it for x by completing the square.

Steps Algebraic
To develop the Quadratic Formula, start with the standard form of a quadratic equation, a times x squared plus b times x plus c equals 0. Remember that a is not equal to zero.
Isolate the variable terms on one side. Isolate the variable terms on one side. The new equation is a times x squared plus b x equals negative c.
Make the coefficient of x2 equal to 1, by
dividing by a.		 Make the leading coefficient 1 by dividing both sides of the equation by a. We now have the quotient a times x squared divided by a plus the product of the quotient b divided by a and x equals negative c divided by a.
Simplify. Simplified, this becomes x squared plus b divided by a times x equals negative c divided by a.
To complete the square, find  \displaystyle\left(\frac{1}{2}\cdot\frac{b}{a}\right)^{2} and add it to both sides of the equation.
\displaystyle\left(\frac{1}{2}\frac{b}{a}\right)^{2}=\frac{b^{2}}{4a^{2}} Complete the square on the left side of the equation. Find the square of one half times the quotient b divided by a which simplifies to b squared divided by the product 4 times a squared. Add this value to both sides of the equation. X squared plus b divided by a times x plus b squared divided by the product 4 a squared equals negative c divided by a plus the quotient b squared divided by the product 4 times a squared.
The left side is a perfect square, factor it. Factor the perfect square trinomial on the left side of the equation. The square of x plus the quotient b divided by 2 a equals the quotient negative c divided by a plus the quotient b squared divided by the product 4 times a squared.
Find the common denominator of the right
side and write equivalent fractions with
the common denominator.		 Find the common denominator of the right side of the equation and write equivalent fractions using the common denominator. Multiply the term negative c divided by a on the right side of the equation by the fraction 4 a divided by 4 a.
Simplify. Rearrange terms on the right side of the equation, and it becomes the square of the sum x plus the quotient b divided by 2 a equals the quotient b squared divided by the product 4 times a squared plus the quotient negative c times 4 a divided by a times 4 a.
Combine to one fraction. Combining to one fraction, the square of the sum x plus the quotient b divided by 2 a equals the quotient of the difference b squared minus 4 a c divided by 4 a squared.
Use the square root property. Use the Square Root Property. X plus the quotient b divded by 2 a equals the positive or negative square root of the quotient of the difference b squared minus 4 a c divided by 4 a squared.
Simplify the radical. Simplify the radical. X plus the quotient b divded by 2 a equals the positive or negative quotient of the square root of the difference b squared minus 4 a c divided by 2 a.
Add  \displaystyle\frac{-b}{2a}  to both sides of the equation.   Add negative b divided by 2 a to both sides of the equation. So x equals negative b divided by the product 2 a plus or minus the quotient of the square root of the difference b squared minus 4 a c divided by 2 a.
Combine the terms on the right side. Combine the terms on the right side of the equation to get the final form of the Quadratic Formula. X equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a.
This equation is the Quadratic Formula.
Quadratic Formula

The solutions to a quadratic equation of the form ax2 + bx + c = 0, where a ≠ 0 are given by the formula:

\displaystyle x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

How to solve quadratic equations using the quadratic formula

  1. Write the quadratic equation in standard form, ax2 + bx + c = 0. Identify the values of a, b, and c.
  2. Write the Quadratic Formula. Then substitute in the values of a, b, and c.
  3. Simplify.
  4. Check the solutions.

Try it!

  1. Solve by using the Quadratic Formula: 2x2 + 9x − 5 = 0.

Solution

Step 1 is to write the quadratic equation in standard form, a times x squared plus b x plus c equals zero, and identify the values a, b, and c. The equation 2 x squared plus 9 x minus 5 equals zero is in standard form. A equals 2, b equals 9, and c equals negative 5.Step 2. Write the quadratic formula. Then substitute the values of a, b, and c. Substitute a equals 2, b equals 9, and c equals negative 5 into the equation x equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. So x equals the quotient negative 9 plus or minus the square root of the difference 9 squared minus the product 4 times 2 times negative 5 divided by the product 2 times 2.In step 3, simplify the fraction and solve for x. x equals the quotient negative 9 plus or minus the square root of the difference 81 minus negative 40 divided by 4. Simplify the radicand. x equals the quotient negative 9 plus or minus the square root of 121 divided by 4. Simplify the square root. x equals the quotient negative 9 plus or minus 11 divided by 4. Separate into two equations. The first equation is x equals the quotient negative 9 plus 11 divided by 4 which simplifies to 2 divided by 4. The first solution is x equals one half. The second equation is x equals the quotient negative 9 minus 11 divided by 4 which simplifies to negative 20 divided by 4. The second solution is x equals negative 5.
The fourth, and final, step is to check the solution. Put each answer into the original equation to check. First, substitute x equals one half into the original equation, 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of one half plus nine times one half minus 5. We need to show that this expression equals 0. Simplify the square. 2 times one fourth plus nine times one half minus 5 equals one half plus 9 halves minus 5, or 10 halves minus 5. 5 minus 5 equals 0, so x equals one half is indeed a solution. Next substitute x = negative 5 into the equation 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of negative 5 plus 9 times negative 5 minus 5. We need to show that this expression equals 0. Simplify the square. 2 times 25 plus nine times negative 5 minus 5 equals 50 minus 45 minus 5, or 0. x equals negative 5 is a solution as well.

2. Solve by using the Quadratic Formula: x2 − 6x = −5.

Solution

Steps Algebraic
Equation The equation x squared minus 6 x equals negative 5 .
Write the equation in standard form by adding
5 to each side.		 Write the equation x squared minus 6 x equals negative 5 in standard form by adding 5 to both sides of the equation. X squared minus 6 x plus 5 equals 0.
This equation is now in standard form.
Shows X squared minus 6 x plus 5 equals 0 with standard form shown above in red (ax squared + bx + c =0.
Identify the values of a, b, c. Identify the values of a, b, and c. The coefficient of x squared is a = 1. The coefficient of x is b equals negative 6. The constant term is c equals 5.
Write the Quadratic Formula. Write the quadratic formula, x equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a.
Then substitute in the values of a, b, c. Then substitute the values for a, b, and c. x equals the quotient of the expression the opposite of negative 6 plus or minus the square root of the difference negative 6 squared minus the product 4 times 1 times 5 divided by the product 2 times 1.
Simplify. Simplify. X equals the quotient of the expression 6 plus or minus the square root of the difference 36 minus 20 divided by 2.This further simplifies to the quotient of 6 plus or minus square root 16 and 2, so x equals the quotient of 6 plus or minus 4 and 2.This further simplifies to the quotient of 6 plus or minus square root 16 and 2, so x equals the quotient of 6 plus or minus 4 and 2.
Rewrite to show two solutions. Rewrite to show two solutions The first is x equals the quotient 6 plus 4 divided by 2, or 10 divided by 2 which equals 5. The second solution is the quotient 6 minus 4 divided by 2, or 2 divided by 2 which equals 1.
Simplify. Rewrite to show two solutions The first is x equals the quotient 6 plus 4 divided by 2, or 10 divided by 2 which equals 5. The second solution is the quotient 6 minus 4 divided by 2, or 2 divided by 2 which equals 1.
Solution Rewrite to show two solutions The first is x equals the quotient 6 plus 4 divided by 2, or 10 divided by 2 which equals 5. The second solution is the quotient 6 minus 4 divided by 2, or 2 divided by 2 which equals 1.
Check:
Check the solutions in the original equation. Substitute x equals 5 into the original equation, x squared minus 6 x plus 5 equals zero to get 5 squared minus 6 times 5 plus 5 on the left side of the equation we must show that this equals 0. Simplifying the expression yields 25 minus 30 plus 5, or 0. So x equals 5 is a solution. Next check x equals 1 in the original equation. X squared minus 6 x plus 5 becomes 1 squared minus 6 times 1 plus 5. We must show this equals 0. 1 minus 6 plus 5 does equal 0, so x = 1 is a solution.

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form.

Try it!

Solve by using the Quadratic Formula: 2x2 + 10x + 11 = 0.

Solution

Steps Algebraic
Equation The equation 2x squared plus 10 x plus 11 equals 0 is already in standard form.
This equation is in standard form. The equation 2x squared plus 10 x plus 11 equals 0 is already in standard form (ax suared + bx + c = 0 is shown above it).
Identify the values of a, b, and c. Identify the values of a, b, and c. The coefficient of x squared is a = 2. The coefficient of x is b equals 10. The constant term is c equals 11.
Write the Quadratic Formula. Write the quadratic formula, x equals the quotient negative b plus or minus the square root of the expression b squared minus 4 a c divided by 2 a.
Then substitute in the values of a, b, and c. Then substitute the values for a, b, and c. x equals the quotient of the difference negative 10 plus or minus the square root of the difference 10 squared minus the product 4 times 2 times 11 divided by the product 2 times 2.
Simplify. Simplify. X equals the quotient of the expression negative 10 plus or minus the square root of the difference 100 minus 88 divided by 4.
This further simplifies to the quotient negative 10 plus or minus square root 12 divided by 4.
Simplify the radical. Simplify the radical. x equals the quotient negative 10 plus or minus 2 times the square root of 3 divided by 2.
Factor out the common factor in the numerator. Factor out the common factor in the numerator. X equals the quotient of 2 times the expression negative 5 plus or minus 2 square root 3 and 4.
Remove the common factors. Remove the common factor to yield x equals the quotient negative 5 plus or minus square root 3 divided by 2.
Rewrite to show two solutions. Rewrite to show two solutions The first solution is x equals the quotient of negative 5 plus square root 3 and 2. The second is x equals the quotient of negative 5 minus square root 3 and 2. Remember to check the solutions in the original equation. We leave that to you!
Check: We leave the check for you!

When we substitute a, b, and c into the Quadratic Formula and the radicand is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.

Try it!

Solve by using the Quadratic Formula: 3p2 + 2p + 9 = 0.

Solution

Steps Algebraic
Equation The equation 3 p squared plus 2 p plus 9 equals 0 is already in standard form.
This equation is in standard form The equation 3 p squared plus 2 p plus 9 equals 0 is already in standard form, which is shown above the equation (ax squared + bx+ c = 0).
Identify the values of a, b, c. Identify the values of a, b, and c. The coefficient of p squared is a = 3. The coefficient of p is b equals 2. The constant term is c equals 9.
Write the Quadratic Formula. Write the quadratic formula, p equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a.
Then substitute in the values of a, b, c. Then substitute the values for a, b, and c. p equals the quotient of the expression negative 2 plus or minus the square root of the difference 4 squared minus the product 4 times 3 times 9 divided by the product 2 times 3.
Simplify. Simplify. P equals the quotient of the expression negative 2 plus or minus the square root of the difference 4 minus 108 divided by 6.
This further simplifies to p equals the quotient negative 2 plus or minus the square root of negative 104 divided by 6.
Simplify the radical using complex numbers. Simplify the radical using complex numbers. p equals the quotient negative 2 plus or minus square root 104 times I divided by 6.
Simplify the radical. Simplify the radical. p equals the quotient negative 2 plus or minus 2 times square root 26 times I divided by 6.
Factor the common factor in the numerator. Factor the common factor in the numerator. P equals the quotient of 2 times the expression negative 1 plus or minus square root 26 times I divided by 6.
Remove the common factors. Remove the common factor to yield p equals the quotient negative 1 plus or minus square root 26 times I divided by 3. Rewrite in standard a plus b I form.
Rewrite in standard a + bi form. Rewrite in standard a plus b I form. P equals negative one third plus or minus square root 26 divided by 3 times I.
Write as two solutions. Write as show two solutions The first solution is p equals negative one third plus square root 26 thirds I. The second is p equals negative one third minus square root 26 thirds I.

Discriminant

When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?

Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the discriminant.

Discriminant

In the Quadratic Formula, \displaystyle x=\frac{-b\pm \sqrt{b^{2} - 4ac}}{2a},  the quantity b2 – 4ac is called the discriminant.

Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations. The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, b2 − 4ac. The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. The table below relates the value of the discriminant to the solutions of a quadratic equation.

When the value under the radical in the Quadratic Formula, the discriminant, is positive, the equation has two real solutions. When the value under the radical in the Quadratic Formula, the discriminant, is zero, the equation has one real solution. When the value under the radical in the Quadratic Formula, the discriminant, is negative, the equation has two complex solutions.

Value of Discriminant Results
b2 − 4ac = 0 One rational solution (double solution)
b2 − 4ac > 0, perfect square Two rational solutions
b2 − 4ac > 0, not a perfect square Two irrational solutions
b2 − 4ac < 0 Two complex solutions

Try it!

  1. Determine the number of solutions to each quadratic equation.

a. 3x2 + 7x − 9 = 0      b.  5n2 + n + 4 = 0      c. 9y2 − 6y + 1 = 0.

Solution

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

a.

Steps Algebraic
Equation 3x2 + 7x − 9 = 0
The equation is in standard form, identify a, b, and c. a = 3, b = 7, c = −9
Write the discriminant. b2 − 4ac
Substitute in the values of a, b, and c. (7)2 − 4 · 3 · (−9)
Simplify. 49 + 108
Solution   157

Since the discriminant is positive, there are 2 real solutions to the equation.

b.

Steps Algebraic
Equation 5n2 + n + 4 = 0
The equation is in standard form, identify a, b, and c. a = 5, b = 1, c = 4
Write the discriminant. b2 − 4ac
Substitute in the values of a, b, and c. (1)2 – 4 · 5 · 4
Simplify. 1 – 80
Solution -79

Since the discriminant is negative, there are 2 complex solutions to the equation.

c.

Steps Algebraic
Equation 9y2 − 6y + 1 = 0
The equation is in standard form, identify a, b, and c. a = 9, b = −6, c = 1
Write the discriminant. b2 − 4ac
Substitute in the values of a, b, and c. (-6)2 – 4 · 9 · 1
Simplify. 36 – 36
Solution 0

Since the discriminant is 0, there is 1 real solution to the equation.

2. For each Equation, compute the determinant and determine the number and type of solution.

  1. x2 + 4x + 4 = 0
  2. 8x2 + 14x + 3 = 0
  3. 3x2 − 5x − 2 = 0
  4. 3x2 − 10x + 15 = 0

Solution:

Calculate the discriminant b2 − 4ac for each equation and state the expected type of solutions.

  1. x2 + 4x + 4 = 0

    b2 − 4ac = (4)2 − 4(1)(4) = 0.

    There will be one rational double solution.

  2. 8x2 + 14x + 3 = 0

    b2 − 4ac = (14)2 − 4(8)(3) = 100.

    As 100 is a perfect square, there will be two rational solutions.

  3. 3x2 − 5x − 2 = 0

    b2 − 4ac = (−5)2 − 4(3)(−2) = 49.

    As 49 is a perfect square, there will be two rational solutions.

  4. 3x2 − 10x + 15 = 0

    b2 − 4ac = (−10)2 − 4(3)(15) = −80.

    There will be two complex solutions.

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is ax2 = k or a(x − h)2 = k we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula. Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.

Try it!

Identify the most appropriate method to use to solve each quadratic equation.

a. 5z2 = 17      b. 4x2 − 12x + 9 = 0      c. 8u2 + 6u = 11.

Solution:

a. 5z2 = 17

Since the equation is in the ax2 = k, the most appropriate method is to use the Square Root Property.

b. 4x2 − 12x + 9 = 0

We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.

c.

Steps Algebraic
Equation 8u2 + 6u = 11
Put the equation in standard form. 8u2 + 6u − 11 = 0

While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.

Access these online resources for additional instruction and practice with using the Quadratic Formula.

 

Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as a2 + b2 = c2, where a and b refer to the legs of a right triangle adjacent to the 90° angle, and c refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

a2 + b2 = c2

where a and b refer to the legs of a right triangle adjacent to the 90° angle, and c refers to the hypotenuse, as shown below.Right triangle with the base labeled: a, the height labeled: b, and the hypotenuse labeled: c

Try it! – Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle

Right triangle with the base labeled: a, the height labeled: 4, and the hypotenuse labeled 12.

Solution

As we have measurements for side b and the hypotenuse, the missing side is a.

Algebraic
a2 + b2 = c2
a2 + (4)2 = (12)2
a2 + 16 = 144
a2 = 128
\displaystyle a=\sqrt{128}
\displaystyle =8\sqrt{2}

Key Concepts

  • Polynomial Equation: A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.
  • Quadratic Equation: An equation of the form ax2 + bx + c = 0 is called a quadratic equation. a, b, and c are real numbers and a ≠ 0 .
  • Zero Product Property: If If a · b = 0, then either a = 0 or b = 0 or both.
  • How to use the Zero Product Property
    1. Set each factor equal to zero.
    2. Solve the linear equations.
    3. Check.
  • How to solve a quadratic equation by factoring.
    1. Write the quadratic equation in standard form, ax2 + bx + c = 0.
    2. Factor the quadratic expression.
    3. Use the Zero Product Property.
    4. Solve the linear equations.
    5. Check. Substitute each solution separately into the original equation.
  • Square Root Property
    • If x2 = k, then \displaystyle x=\sqrt{k} or \displaystyle x=-\sqrt{k} or \displaystyle x=\pm\sqrt{k}.
  • How to solve a quadratic equation using the square root property.
    1. Isolate the quadratic term and make its coefficient one.
    2. Use Square Root Property.
    3. Simplify the radical.
    4. Check the solutions.
  • Quadratic Equation: The solutions to a quadratic equation of the form ax2 + bx + c = 0, where a ≠ 0 are given by the formula: \displaystyle x=\frac{-b\pm \sqrt{b^{2} - 4ac}}{2a}
  • Using the Discriminant, b2 − 4ac, to Determine the Number and Type of Solutions of a Quadratic Equation
    • For a quadratic equation of the form ax2 + bx + c = 0, a ≠ 0,
      • If b2 − 4ac > 0, the equation has 2 real solutions.
      • if b2 − 4ac = 0, the equation has 1 real solution.
      • if b2 − 4ac < 0, the equation has 2 complex solutions.
definition

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