College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Mid-Chapter Check Point - Page 165: 6


$$x_{1} = \frac{6\sqrt 5}{5}$$ $$x_{2} = \frac{-6\sqrt 5}{5}$$

Work Step by Step

We have $$5x^2 + 1 = 37$$ We are going to take numbers to the right side of the equation and $x$ terms to the left side $$5x^2 = 37 - 1$$ Simplifying $$5x^2 = 36$$ Taking the $5$ which is multiplying the $x$ to the other side dividing $$x^2 = \frac{36}{5}$$ And now taking the power squared to the other side as square root $$x = \sqrt {\frac{36}{5}}$$ We can distribute square root on both parts of the fraction $$x = \frac{\sqrt {36}}{\sqrt 5}$$ As we know square roots have two values one positive and the other one negative. $5$ does not has exact square root but $36$ has. So for the first value of $x$ we have $$x_{1} = \frac{6}{\sqrt 5}$$ And for the second value of $x$ we have $$x_{2} = \frac{-6}{\sqrt 5}$$ We can rationalize both values multiplying by $\sqrt 5$ both the numerator and the denominator of each one and we get $$x_{1} = \frac{6*\sqrt 5}{\sqrt 5*\sqrt 5} = \frac{6\sqrt 5}{5}$$ $$x_{2} = \frac{-6*\sqrt 5}{\sqrt 5*\sqrt 5} = \frac{-6\sqrt 5}{5}$$
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