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Quadratic Equation: Solving by the square root method This method can be used if the quadratic equation can be put in the form au2 + bu + c = 0, where b = 0 and u is an algebraic expression. In other words, there is no "bu" term. Example 1: Solve 3x2 + 36 = 0. First, isolate x2. 3x2 = - 36 x2 = - 12 Next, take the square root of each side and place a "" symbol in front of the root on the right side. Table of Contents x2 12 x 12 Quadratic Equation: Solving by the square root method Last, simplify the radical expression. x 12 x 12i x 2 3i Try to solve: 2x2 – 80 = 0. The solutions set is: { 2 10 }. Notes: The "" symbol is placed in front of the right side because x 2 x . For example, x2 = 9 becomes x 2 9 or x 9 which has two solutions, 3. Table of Contents Slide 2 Quadratic Equation: Solving by the square root method Example 2: Solve (3x – 5)2 + 36 = 0. Note: this has the form au2 + bu + c = 0, where b = 0 and u is an algebraic expression (u = 3x – 5). So first, isolate u2 = (3x – 5)2. Next, take the square root of each side and place a "" symbol in front of the root (3x – 5)2 = - 36 3x 5 36 2 3x 5 36 on the right side. Next, solve for x. 3x 5 36 5 36 x 3 Table of Contents Slide 3 Quadratic Equation: Solving by the square root method 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). 5 36 x , 3 5 6 x i, 3 3 5 36i x , 3 5 x 2i 3 5 6i x , 3 Try to solve: Solve (4x + 1)2 = 24. 1 2 6 The solutions set is: . 4 Table of Contents Slide 4 Quadratic Equation: Solving by the square root method Table of Contents