Solve a quadratic equation EXAMPLE 1 Solve a quadratic equation Solve 2x2 + 11 = –37. 2x2 + 11 = –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 6 and –2i 6 .

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 = 13 + 3i 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.

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.

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.

Multiply complex numbers EXAMPLE 4 Multiply complex numbers b. (9 – 2i)(– 4 + 7i) = – 36 + 63i + 8i – 14i2 Multiply using FOIL. = – 36 + 71i – 14(– 1) Simplify and use i2 = – 1 . = – 36 + 71i + 14 Simplify. = –22 + 71i Write in standard form.

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. – 13 + 33i 17 = Simplify.

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

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 4 + 5 = 9 and 4 –5 = – 1. ANSWER

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

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.

( ) 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 = – 4 + 36 Add –12 2 ( ) = (–6) 36 to each side. (x – 6)2 = 32 Write left side as a binomial squared. x – 6 = + 32 Take square roots of each side. x = 6 + 32 Solve for x. x = 6 + 4 2 Simplify: 32 = 16 2 4 The solutions are 6 + 4 and 6 – 4 2 ANSWER

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 0.34 6 – 4 2 and 11.66 6 + 4 2

( ) 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 = –2 + –3 Solve for x. x = –2 + i 3 Write in terms of the imaginary unit i.

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

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

( ) 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 = 24 + 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.

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

( ) 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