Solve an equation with two real solutions EXAMPLE 1 Solve an equation with two real solutions Solve x2 + 3x = 2. x2 + 3x = 2 Write original equation. x2 + 3x – 2 = 0 Write in standard form. x = – b + b2 – 4ac 2a Quadratic formula x = – 3 + 32 – 4(1)(–2) 2(1) a = 1, b = 3, c = –2 x = – 3 + 17 2 Simplify. The solutions are x = –3 + 17 2 0.56 and –3 – –3.56. ANSWER

EXAMPLE 1 Solve an equation with two real solutions CHECK Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about –3.56. 

Solve an equation with one real solutions EXAMPLE 2 Solve an equation with one real solutions Solve 25x2 – 18x = 12x – 9. 25x2 – 18x = 12x – 9. Write original equation. 25x2 – 30x + 9 = 0. Write in standard form. x = 30 + (–30)2– 4(25)(9) 2(25) a = 25, b = –30, c = 9 x = 30 + 0 50 Simplify. x = 3 5 Simplify. 3 5 The solution is ANSWER

EXAMPLE 2 Solve an equation with one real solutions CHECK Graph y = –5x2 – 30x + 9 and note that the only x-intercept is 0.6 = . 3  5

Solve an equation with imaginary solutions EXAMPLE 3 Solve an equation with imaginary solutions Solve –x2 + 4x = 5. –x2 + 4x = 5 Write original equation. –x2 + 4x – 5 = 0. Write in standard form. x = –4 + 42 – 4(–1)(–5) 2(–1) a = –1, b = 4, c = –5 x = –4 + –4 –2 Simplify. –4 + 2i x = –2 Rewrite using the imaginary unit i. x = 2 + i Simplify. The solution is 2 + i and 2 – i. ANSWER

EXAMPLE 3 Solve an equation with imaginary solutions CHECK Graph y = 2x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + i is shown. –(2 + i)2 + 4(2 + i) = 5 ? –3 – 4i + 8 + 4i = 5 ? 5 = 5 

GUIDED PRACTICE for Examples 1, 2, and 3 Use the quadratic formula to solve the equation. x2 = 6x – 4 ANSWER 3 + 5 4x2 – 10x = 2x – 9 1 2 ANSWER 1 7x – 5x2 – 4 = 2x + 3 115 5 + i 10 ANSWER