Lesson 5–5

Example 1  Equation with Rational Roots
Solve 2x2 – 36x + 162 = 32 by using the Square Root Property.

 2x2 – 36x + 162 = 32	Original equation
2(x2 – 18x + 81) = 2(16)	Factor out the GCF.
     x2 – 18x + 81 = 16	Divide each side by 2.
   (x – 9)2 = 16	Factor the perfect trinomial square.
    x – 9 = ±	Square Root Property
    x – 9 = ±4	 = 4
x = 9 ± 4	Add 9 to each side.
x = 9 + 4       or    x = 9 – 4	Write as two equations.
x = 13                  x = 5	Solve each equation.

The solution set is {5, 13}. You can check this result by using factoring to solve the original equation.

Example 2  Equation with Irrational Roots
Solve x2 + 10x + 25 = 108 by using the Square Root Property.

x2 + 10x + 25 = 108	Original equation
         (x + 5)2 = 108	Factor the perfect square trinomial.
   x + 5 = ±	Square Root Property
         x = –5 ±6	Add –5 to each side;  = 6
x = –5 + 6   or   x = –5 – 6 	Write as two equations.
x ≈ 5.4                     x ≈ –15.4	Use a calculator.

The exact solutions of this equation are –5 – 6 and –5 + 6. The approximate solutions are –15.4 and 5.4. Check these results by finding and graphing the related quadratic function.

x2 + 10x + 25 = 108	Original equation
x2 + 10x – 83 = 0	Subtract 108 from each side.
y = x2 + 10x – 83	Related quadratic function.

CHECK:  Use the ZERO function of a graphing calculator. The approximate zeros of the related function are –15.4 and 5.4.

Example 3  Complete the Square
Find the value of c that makes x2 – x + c a perfect square. Then write the trinomial as a perfect square.

Step 1	Find one half of .	 •  = 
Step 2	Square the result of Step 1.	= 
Step 3	Add the result of Step 2 to x2 – x	x2 – x + 

The trinomial x2 – x +  can be written as . 

Example 4  Solve an Equation by Completing the Square
Solve x2 – 7x – 44 = 0 by completing the square.

x2 – 7x – 44 = 0	Notice that x2 – 7x – 44 is not a perfect square.
x2 – 7x = 44 	Rewrite so the left side is of the form x2 + bx.
x2 – 7x +  = 44 + 	Since  = , add  to each side.
       = 	Write the left side as a perfect square by factoring.
x –  = 	Square Root Property
x –  = 	 = 
x =  	Add  to each side.
x =  +    or   x =  – 	Write as two equations.
x = 11                  x = –4	The solution set is {-4, 11}.

You can check this result by using factoring to solve the original equation.

Example 5  Equation with a ≠ 1
Solve 3x2 + 4x – 7 = 0 by completing the square.
3x2 + 4x – 7 = 0	Notice that 3x2 + 4x – 7 is not a perfect square.
x2 + x –  = 0	Divide by the coefficient of the quadratic term, 3.
x2 + x = 	Add  to each side.
x2 + x +  =  + 	Since  = , add  to each side.
 = 	Write the left side as a perfect square by factoring.
	Simplify the right side.
x +  = ±	Square Root Property
       x = – ±	Add – to each side.
x = – +      or    x = – – 	Write as two equations.
x = 1                        x = –	The solution set is .

Example 6  Equation with Complex Solutions
Solve 4x2 – 2x + 7 = 0 by completing the square.

4x2 – 2x + 7 = 0	Notice that 4x2 – 2x + 7 is not a perfect square.
x2 – x +  = 0	Divide by the coefficient of the quadratic term, 4.
 x2 – x = –	Rewrite so the left side is of the form x2 + bx.
x2 – x +  = – + 	Since  = , add  to each side.
   = –	Write the left side as a perfect square by factoring.
	Simplify the right side.
x –  = 	Square Root Property
x –  = ±	
       x =  ±	Add  to each side.

The solution set is . Notice that these are imaginary solutions.

CHECK: A graph of the related function shows that the equation has no real solutions since the graph has no x–intercepts. Imaginary solutions must be checked algebraically by substituting them back in the original equation.