1. Solve a quadratic equation
EXAMPLE 1
Solve a quadratic equation
Solve 2x = –37.
2x = –37
Write original equation.
2x2 = – 48
Subtract 11 from each side.
x2 = – 24
Divide each side by 2.
x = + – 24
Take square roots of each side.
x = + i 24
Write in terms of i.
x = + 2i 6
Simplify radical.
ANSWER
The solutions are 2i and –2i

2. Add and subtract complex numbers
EXAMPLE 2
Add and subtract complex numbers
Write the expression as a complex number in standard form.
a. (8 – i) + (5 + 4i)
b. (7 – 6i) – (3 – 6i)
c. 10 – (6 + 7i) + 4i
SOLUTION
a. (8 – i) + (5 + 4i) =
Definition of complex addition
(8 + 5) + (–1 + 4)i
= i
Write in standard form.
b. (7 – 6i) – (3 – 6i) =
Definition of complex subtraction
(7 – 3) + (–6 + 6)i
= 4 + 0i
Simplify.
= 4
Write in standard form.

3. Add and subtract complex numbers
EXAMPLE 2
Add and subtract complex numbers
c. 10 – (6 + 7i) + 4i =
Definition of complex subtraction
[(10 – 6) – 7i] + 4i
= (4 – 7i) + 4i
Simplify.
= 4 + (– 7 + 4)i
Definition of complex addition
= 4 – 3i
Write in standard form.

4. Multiply complex numbers
EXAMPLE 4
Multiply complex numbers
Write the expression as a complex number in standard form.
a. 4i(–6 + i)
b. (9 – 2i)(–4 + 7i)
SOLUTION
a. 4i(– 6 + i) = – 24i + 4i2
Distributive property
= – 24i + 4(– 1)
Use i2 = –1.
= – 24i – 4
Simplify.
= – 4 – 24i
Write in standard form.

5. Multiply complex numbers
EXAMPLE 4
Multiply complex numbers
b. (9 – 2i)(– 4 + 7i)
= – i + 8i – 14i2
Multiply using FOIL.
= – i – 14(– 1)
Simplify and use i2 = – 1 .
= – i + 14
Simplify.
= – i
Write in standard form.

6. Divide complex numbers
EXAMPLE 5
Divide complex numbers
Write the quotient in standard form.
7 + 5i
1  4i
7 + 5i
1 – 4i =
1 + 4i
Multiply numerator and denominator by 1 + 4i, the complex conjugate of 1 – 4i.
7 + 28i + 5i + 20i2
1 + 4i – 4i – 16i2 =
Multiply using FOIL.
7 + 33i + 20(– 1)
1 – 16(– 1) =
Simplify and use i2 = 1.
– i 17 =
Simplify.

7. Divide complex numbers
EXAMPLE 5
Divide complex numbers 13 17 – = + 33 i
Write in standard form.

8. Solve a quadratic equation by finding square roots
EXAMPLE 1
Solve a quadratic equation by finding square roots
Solve x2 – 8x + 16 = 25.
x2 – 8x + 16 = 25
Write original equation.
(x – 4)2 = 25
Write left side as a binomial squared.
x – 4 = +5
Take square roots of each side.
x = 4 + 5
Solve for x.
The solutions are = 9 and 4 –5 = – 1.
ANSWER

9. EXAMPLE 2
Make a perfect square trinomial
Find the value of c that makes x2 + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial.
SOLUTION
STEP 1 16 2 = 8
Find half the coefficient of x.
STEP 2
Square the result of Step 1.
82 = 64
STEP 3
Replace c with the result of Step 2.
x2 + 16x + 64

10. EXAMPLE 2
Make a perfect square trinomial
ANSWER
The trinomial x2 + 16x + c is a perfect square when c = 64. Then x2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.

11. ( ) EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1
Solve x2 – 12x + 4 = 0 by completing the square.
x2 – 12x + 4 = 0
Write original equation.
x2 – 12x = – 4
Write left side in the form x2 + bx.
x2 – 12x + 36 = –
Add
–12 2 ( ) =
(–6) 36
to each side.
(x – 6)2 = 32
Write left side as a binomial squared.
x – 6 =
Take square roots of each side.
x =
Solve for x.
x =
Simplify: 32 = 16 2 4
The solutions are 6 + 4
and 6 – 4 2
ANSWER

12. EXAMPLE 3
Solve ax2 + bx + c = 0 when a = 1
CHECK
You can use algebra or a graph.
Algebra Substitute each solution in the original equation to verify that it is correct.
Graph Use a graphing calculator to graph
y = x2 – 12x + 4. The x-intercepts are about – 4 2 and

13. ( ) EXAMPLE 4 Solve ax2 + bx + c = 0 when a = 1
Solve 2x2 + 8x + 14 = 0 by completing the square.
2x2 + 8x + 14 = 0
Write original equation.
x2 + 4x + 7 = 0
Divide each side by the coefficient of x2.
x2 + 4x = – 7
Write left side in the form x2 + bx.
Add 4 2 ( ) =
to each side.
x2 – 4x + 4 = – 7 + 4
(x + 2)2 = –3
Write left side as a binomial squared.
x + 2 = –3
Take square roots of each side.
x = – –3
Solve for x.
x = –2 + i 3
Write in terms of the imaginary unit i.

14. EXAMPLE 4
Solve ax2 + bx + c = 0 when a = 1
The solutions are –2 + i 3
and – 2 – i
3 .
ANSWER

15. EXAMPLE 5
Standardized Test Practice
SOLUTION
Use the formula for the area of a rectangle to write an equation.

16. ( ) EXAMPLE 5 Standardized Test Practice 3x(x + 2) = 72 3x2 + 6x = 72
Length Width = Area
3x2 + 6x = 72
Distributive property
x2 + 2x = 24
Divide each side by the coefficient of x2.
Add 2 ( ) = 1
to each side.
x2 – 2x + 1 =
(x + 1)2 = 25
Write left side as a binomial squared.
x + 1 = + 5
Take square roots of each side.
x = –1 + 5
Solve for x.

17. EXAMPLE 5
Standardized Test Practice
So, x = –1 + 5 = 4 or x = – 1 – 5 = – 6. You can reject x = – 6 because the side lengths would be – 18 and – 4, and side lengths cannot be negative.
The value of x is 4. The correct answer is B.
ANSWER

18. ( ) EXAMPLE 6 Write a quadratic function in vertex form
Write y = x2 – 10x + 22 in vertex form. Then identify the vertex.
y = x2 – 10x + 22
Write original function.
y + ? = (x2 –10x + ? ) + 22
Prepare to complete the square.
Add
–10 2 ( ) =
(–5) 25
to each side.
y + 25 = (x2 – 10x + 25) + 22
y + 25 = (x – 5)2 + 22
Write x2 – 10x + 25 as a binomial squared.
y = (x – 5)2 – 3
Solve for y.
The vertex form of the function is y = (x – 5)2 – 3. The vertex is (5, – 3).
ANSWER