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Published byEdwina Veronica Henderson Modified over 6 years ago

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Table of Contents Example 1: Solve 3x 2 + 36 = 0. Quadratic Equation: Solving by the square root method This method can be used if the quadratic equation can be put in the form au 2 + bu + c = 0, where b = 0 and u is an algebraic expression. In other words, there is no "bu" term. First, isolate x 2.3x 2 = - 36 Next, take the square root of each side and place a " " symbol in front of the root on the right side. x 2 = - 12

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Table of Contents Try to solve: 2x 2 – 80 = 0. Quadratic Equation: Solving by the square root method Slide 2 Last, simplify the radical expression. The solutions set is: Notes: The " " symbol is placed in front of the right side because For example, x 2 = 9 becomes or which has two solutions, 3.

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Table of Contents Quadratic Equation: Solving by the square root method Slide 3 Example 2: Solve (3x – 5) 2 + 36 = 0. Note: this has the form au 2 + bu + c = 0, where b = 0 and u is an algebraic expression (u = 3x – 5). So first, isolate u 2 = (3x – 5) 2. (3x – 5) 2 = - 36 Next, take the square root of each side and place a " " symbol in front of the root on the right side. Next, solve for x.

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Table of Contents Try to solve: Solve (4x + 1) 2 = 24. Quadratic Equation: Solving by the square root method Slide 4 Last, simplify the radical expression. If a fraction results, simplify it. If nonreal solutions result it is customary to write them in standard form (a + bi). The solutions set is:

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Table of Contents Quadratic Equation: Solving by the square root method

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