Explanation Given Information: The provided equation is p 4 − 15 p 2 = − 54 . Formula used: The formula for solving a Quadratic Equation a x 2 + b x + c = 0 by completing the square and applying square root property, Step 1. Divide both sides by a to make the leading coefficient 1. Step 2. Isolate the variable terms on one side of the equation. Step 3. Complete the square, Add the square of one half the linear term coefficient to both sides, ( 1 2 b ) 2 . Factor the resulting perfect square trinomial. Step 4. Apply the square root property and solve for x . The square root property, For any real number, k , if x 2 = k , then x = k or x = − k . Calculations: Consider the provided equation, p 4 − 15 p 2 = − 54 Consider, u = p 2 , Therefore, u 2 − 15 u + 54 = 0 Now, factorize the given equation as follows: u 2 − 15 u + 54 = 0 u 2 − 15 u = − 54 u 2 − 13 u + 225 4 = − 54 + 225 4 ( u − 15 2 ) 2 = 9 4 Now using square root property, u − 15 2 = ± 9 4 u − 15 2 = ± 3 2 Solving further as shown below, u = 3 2 + 15 2 = 3 + 15 2 = 18 2 = 9 And further, u = − 3 2 + 15 2 = − 3 + 15 2 = 12 2 = 6 Hence, u = 9 , 6

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ISBN: 9781264667765

Author: Miller

Publisher: MCG

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Chapter 11, Problem 16T

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Chapter 11 Solutions

ALEKS 360 ACCESS: BEG.+INTERM ALG.

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The solution of the equation p 4 − 15 p 2 = − 54 .

Solution Summary: The author explains the formula for solving a Quadratic Equation by completing the square and applying square root property.

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