Explanation Definition used: “Standard q uadratic equation: “An equation in which the variable appears to the second power, but to no higher power, is called a quadratic equation . A standard quadratic equation can be expresses as, a x 2 ︸ + b x ︸ + c ︸ = 0 ↓ ↓ ↓ [ x 2 -term ] [ x -term ] [ Constant term ] That is every quadratic equation must have x 2 -term, but the x -term or constant tem may be missing.” Procedure used: 1. Identify the values of a , b and c by comparing the given equation with the standard quadratic equation a x 2 + b x + c = 0 . 2. Substitute the values of a , b and c into the quadratic equation formula x = − b ± b 2 − 4 a c 2 a . 3. Simplify the equation for x . 4. Check the solution by back substituting the values of x in the original equation. Calculation: The given quadratic equation is 0.4 x 2 + 0.6 x − 0.8 = 0 . Step 1 : Identify the values of a , b and c by comparing the given equation with the standard quadratic equation a x 2 + b x + c = 0 . 0 .4 x 2 ︸ + 0 .6 x ︸ + − 0.8 ︸ = 0 ↓ ↓ ↓ [ a = 0

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ISBN: 2818440003295

Author: CARMAN

Publisher: PEARSON CP

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Chapter 11.2, Problem 9BE

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To solve: The equation 0.4x2+0.6x−0.8=0.

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Chapter 11.2, Problem 8BE

Chapter 11.2, Problem 10BE

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The equation 0.4 x 2 + 0.6 x − 0.8 = 0 .

Solution Summary: The author explains the solution of the equation 0.4x2+0.6x-0.8=0 by comparing the given equation with the standard quadratic equation.

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To solve: The equation 0.4x2+0.6x−0.8=0.

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Chapter 11 Solutions