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4.8 Use the Quadratic Formula and the Discriminant General Equation Quadratic Formula

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EXAMPLE 1 Solve an equation with two real solutions Solve x 2 + 3x = 2. x 2 + 3x = 2 Write original equation. x 2 + 3x – 2 = 0 Write in standard form. x = – b + b 2 – 4ac 2a2a Quadratic formula x = – – 4(1)( –2) 2(1) a = 1, b = 3, c = –2 Simplify. x = – 3 +– The solutions are x = – and x = –3 – 17 2 –3.56. ANSWER

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

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

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

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

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EXAMPLE 3 Solve an equation with imaginary solutions CHECK Graph y = 2x 2 + 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 i = 5 ? 5 = 5

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

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EXAMPLE 4 Use the discriminant Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. a. x 2 – 8x + 17 = 0 b. x 2 – 8x + 16 = 0 c. x 2 – 8x + 15 = 0 SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac x = – b+ b 2 – 4ac 2ac a. x 2 – 8x + 17 = 0(–8) 2 – 4(1)(17) = –4 Two imaginary: 4 + i b. x 2 – 8x + 16 = 0(–8) 2 – 4(1)(16) = 0 One real: 4 b. x 2 – 8x + 15 = 0(–8) 2 – 4(1)(15) = 0 Two real: 3,5

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GUIDED PRACTICE for Example 4 Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. 4. 2x 2 + 4x – 4 = 0 48 ; Two real solutions 5. 0 ; One real solution 3x x + 12 = x 2 = 9x – 11 –271 ; Two imaginary solutionsANSWER

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7. GUIDED PRACTICE for Example 4 7x 2 – 2x = ; Two real solutions 8. 4x 2 + 3x + 12 = 3 – 3x –108 ; Two imaginary solutions 9. 3x – 5x = 6 – 7x 0 ; One real solutionANSWER

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EXAMPLE 1 Graph a quadratic inequality Graph y > x 2 + 3x – 4. SOLUTION STEP 1 Graph y = x 2 + 3x – 4. Because the inequality symbol is >, make the parabola dashed. Test a point inside the parabola, such as (0, 0). STEP 2 y > x 2 + 3x – 4 0 > (0) – 4 ? 0 > – 4

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EXAMPLE 1 Graph a quadratic inequality So, (0, 0) is a solution of the inequality. STEP 3 Shade the region inside the parabola.

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EXAMPLE 3 Graph a system of quadratic inequalities Graph the system of quadratic inequalities. y < –x Inequality 1 y > x 2 – 2x – 3 Inequality 2 SOLUTION STEP 1 Graph y ≤ –x The graph is the red region inside and including the parabola y = –x

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EXAMPLE 3 Graph a system of quadratic inequalities STEP 2 Graph y > x 2 – 2x – 3. The graph is the blue region inside (but not including) the parabola y = x 2 – 2x – 3. Identify the purple region where the two graphs overlap. This region is the graph of the system. STEP 3

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GUIDED PRACTICE for Examples 1, 2, and 3 Graph the inequality. 1. y > x 2 + 2x – 8y < 2x 2 – 3x

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GUIDED PRACTICE for Examples 1, 2, and 3 Graph the inequality. y < –x 2 + 4x Graph the system of inequalities consisting of y ≥ x 2 and y < –x

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