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Algebra 1 Section 12.6
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Solving Quadratic Equations
Choose the method that is most appropriate for the given problem. Square root property Factoring Quadratic formula
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Solving Quadratic Equations
Completing the square is usually not used since one of the other techniques is usually easier. However, it is valuable in other applications you will use later.
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Solving Quadratic Equations
Write the quadratic equation in the form ax2 + bx + c = 0. If b = 0, isolate x2 and apply the Square Root Property.
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Solving Quadratic Equations
If the polynomial factors easily, solve the equation by factoring. Otherwise, use the quadratic formula: x = -b ± b2 – 4ac 2a
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The Discriminant The radicand in the quadratic formula, b2 – 4ac, determines the number and type of solutions to the equation. This value, D = b2 – 4ac, is called the discriminant.
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The Discriminant Number and type of solutions D = b2 – 4ac
D > 0 and a perfect square 2 rational D > 0 but not a perfect square 2 irrational
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Number and type of solutions
The Discriminant Number and type of solutions D = b2 – 4ac D = 0 1 rational no real solutions D < 0
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Example 1a x2 + 5x – 7 = 0 a = 1, b = 5, c = -7 D = b2 – 4ac
= 52 – 4(1)(-7) = 25 – (-28) D = 53 There are 2 irrational solutions.
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Example 1b x2 = 156 – x x2 + x – 156 = 0 a = 1, b = 1, c = -156
D = b2 – 4ac = 12 – 4(1)(-156) = 625 There are 2 rational solutions.
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Example 1c 3x2 + 5x = -9 3x2 + 5x + 9 = 0 a = 3, b = 5, c = 9
D = b2 – 4ac = 52 – 4(3)(9) = -83 There are no real solutions.
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Example 2a If the solutions are x = 5 and x = -7, then we know...
...and using the Multiplication Property of Equality, we get: x2 + 2x – 35 = 0 (x – 5)(x + 7) = 0 • 0
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Example 2b x = ⅔ and x = -¼ Multiply to clear the fractions:
(3x – 2)(4x + 1) = 0 • 0 12x2 – 5x – 2 = 0
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Example 3 Let x = hypotenuse length x – 3 = length of one leg
x – 5 = length of other leg Since a2 + b2 = c2, (x – 5)2 + (x – 3)2 = x2
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The solutions will be irrational.
Example 3 (x – 5)2 + (x – 3)2 = x2 x2 – 10x x2 – 6x + 9 = x2 x2 – 16x + 34 = 0 D = b2 – 4ac D = (-16)2 – 4(1)(34) = 120 The solutions will be irrational.
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Example 3 x2 – 16x + 34 = 0 D = 120 8 1 x = 16 ± 120 2 = 16 ± 2 1 x = 8 ± 30
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Example 3 x = 8 ± 30 x = ≈ ft x = 8 – 30 ≈ 2.52 ft
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Example 3 Let x = hypotenuse length x – 3 = length of one leg
x – 5 = length of other leg Hypotenuse ≈ ft Leg1 = x – 5 ≈ 8.48 ft Leg2 = x – 3 ≈ ft
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Homework: pp
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