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Published byBeryl Hopkins Modified over 9 years ago
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Warm-Up Solve each equation by factoring. 1) x 2 + 13x + 36 = 02) 2x 2 + 5x = 12
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Simplifying Radical Expressions Simplifying Radical Expressions Objectives: To simplify radical expressions
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Perfect Squares 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 324 400 625 289
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= 2 = 4 = 5 = 10 = 12
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= = = = = = = = = = Perfect Square Factor * Other Factor LEAVE IN RADICAL FORM
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Product Property for Radicals For any nonnegative real numbers a and b, So,
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Example 1 Simplify.
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The Quadratic Formula The Quadratic Formula
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If ax 2 + bx + c = 0, a = 0, then the quadratic formula gives the solutions of the quadratic equation.
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Example 1 Solve using the quadratic formula. 3x 2 – 7x + 2 = 0
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Example 1 Solve using the quadratic formula. 3x 2 – 7x + 2 = 0
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Example 2 Solve using the quadratic formula. -x 2 + x + 1 = 0
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Example 2 Solve using the quadratic formula. -x 2 + x + 1 = 0
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Practice 1) 2x 2 – 4x = 5 Solve using the quadratic formula. 2) x 2 + 5x = -3 Approximate the solutions to the nearest tenth.
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Practice 1) 6x 2 – 6 – 5x Use the quadratic formula to find the roots of each polynomial. 2) 6x 2 – 6x - 5
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The Discriminant The expression under the radical, b 2 – 4ac, is called the discriminant. We can use the discriminant to determine the number of solutions b 2 – 4ac is positive: two real number solutions b 2 – 4ac is zero: one real number solution b 2 – 4ac is negative: no real number solutions
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Example 1 Determine the number of real solutions, then solve using the quadratic formula. 4x 2 – 7x + 2 = 0 b 2 – 4ac = 49 – (4)(4)(2) b 2 – 4ac = 17 two real solutions
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Practice 1) x 2 + 5x = -8 2) 4x 2 = 8x - 4 Determine the number of real solutions, then solve using the quadratic formula. 3) 4x 2 + 4x = 15
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Practice 1) y = 3x 2 - 5x + 1 2) f(x) = x 2 – 3x + 7 Use the discriminant to determine whether the graph of each quadratic function intersects the x-axis in 0,1, or 2 points. 3) y = x 2 – 12x + 36
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