## College Algebra (6th Edition)

Published by Pearson

# Chapter 1 - Mid-Chapter Check Point - Page 165: 2

#### Answer

$$x_{1} = \frac{7}{5}$$ $$x_{2} = -1$$

#### Work Step by Step

We have $$5x^2 - 2x = 7$$ As we see we have a quadratic term, a linear term and a constant so we are going to put all the terms together on the left side of the equation to equalize it to zero $$5x^2 - 2x - 7 = 0$$ Using the quadratic formula where $a = 5$, $b = -2$ and $c = -7$ $$x_{1,2} =\frac{-b ±\sqrt {b^2 - 4ac}}{2a} = \frac{-(-2) ±\sqrt {(-2)^2 - 4(5)(-7)}}{2(5)}$$ Evaluating each operation taking in count the sings law $$x_{1,2} =\frac{2 ±\sqrt {4 + 140}}{10} =\frac{2 ±\sqrt {144}}{10} = \frac{2 ±12}{10}$$ For the first value of $x$ we get $$x_{1} = \frac{2+12}{10} = \frac{14}{10} = \frac{7}{5}$$ And for the second value of $x$ we get $$x_{2} = \frac{2-12}{10} = \frac{-10}{10} = -1$$

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