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Example 1: Solve 3 x 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. Example 1: Solve 3 x 2 + 36 = 0.

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Example 1: Solve 3 x 2 + 36 = 0.

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  1. 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.

  2. The solutions set is: Notes: The "" symbol is placed in front of the right side because For example, x2 = 9 becomes or which has two solutions,  3. Quadratic Equation: Solving by the square root method Last, simplify the radical expression. Try to solve: 2x2 – 80 = 0. Slide 2

  3. 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. (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. Slide 3

  4. The solutions set is: 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). Try to solve: Solve (4x + 1)2 = 24. Slide 4

  5. Quadratic Equation: Solving by the square root method END OF PRESENTATION Click to rerun the slideshow.

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