# Warm Up #4 1. Write 15x2 + 6x = 14x2 – 12 in standard form. ANSWER

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Warm Up #4 1. Write 15x2 + 6x = 14x2 – 12 in standard form. ANSWER
2. Evaluate b2 – 4ac when a = 3, b = –6, and c = 5. ANSWER –24

4.6 (multiples of 3)

When solving quadratic equations, we’re looking for the x values where the graph crosses the x axis
One of the methods we use to solve quadratic equations is called the Quadratic Formula Using the a, b, and c from ax2 + bx + c = 0 Must be equal to zero

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 = – – 4(1)(–2) 2(1) a = 1, b = 3, c = –2 x = – 3 + 17 2 Simplify. or

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)2– 4(25)(9) 2(25) a = 25, b = –30, c = 9 x = 50 Simplify. x = 3 5 Simplify. 3 5 The solution is ANSWER

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 = – 42 – 4(–1)(–5) 2(–1) a = –1, b = 4, c = –5 x = – –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

GUIDED PRACTICE for Examples 1, 2, and 3 Use the quadratic formula to solve the equation. x = – b + b2 – 4ac 2a x2 = 6x – 4 x2 – 6x + 4 = 0 a = 1 b = -6 c = 4

GUIDED PRACTICE for Examples 1, 2, and 3 Use the quadratic formula to solve the equation. 4x2 – 10x = 2x – 9 x = – b + b2 – 4ac 2a 4x2 – 12x + 9 = 0 a = 4 b = -12 c = 9

EXAMPLE 4 Use the discriminant If the quadratic equation is in the standard form ax2 + bx + c = 0 The discriminant can be found using b2 – 4ac If b2 – 4ac < There are Two Imaginary solutions If b2 – 4ac = There is One Real solution If b2 – 4ac > There are Two Real solutions Discriminant Equation b2 – 4ac Solution(s) a. x2 – 8x + 17 = 0 ( –8)2 – 4(1)(17) = –4 Two imaginary b. x2 – 8x + 16 = 0 (–8)2 – 4(1)(16) = 0 One real b. x2 – 8x + 15 = 0 (–8)2 – 4(1)(15) = 4 Two real

GUIDED PRACTICE for Example 4 Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. 3x2 + 12x + 12 = 0 2x2 + 4x – 4 = 0 b2 – 4ac b2 – 4ac (4)2 – 4(2)(-4) = 48 (12)2 – 4(3)(12) = 0 Two Real solutions One Real solution 8x2 = 9x – 11 7. 4x2 + 3x + 12 = 3 – 3x 8x2 – 9x + 11 = 0 4x2 + 6x + 9 = 0 b2 – 4ac b2 – 4ac (6)2 – 4(4)(9) = -108 (-9)2 – 4(8)(11) = -271 Two Imaginary solutions Two Imaginary solutions

Classwork Assignment:
WS 4.8 (1-27 odd)

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