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Q. 105 E

Expert-verified

Intermediate Algebra

Found in: Page 885

Book edition OER 2017

Author(s) OPENSTAX

Pages 1346 pages

ISBN 9780998625720

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Short Answer

In the following exercises, solve by completing the square.
$2 x^{2} + 7 x - 15 = 0$

The solution is $x = - 5, x = \frac{3}{2}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

Equation: $2 x^{2} + 7 x - 15 = 0$

Step 2. Calculation

We have,

$2 x^{2} + 7 x - 15 = 0$

$2 x^{2} + 7 x - 15 + 15 = 0 + 15$

$\frac{2 x^{2} + 7 x}{2} = \frac{15}{2}$

role="math" localid="1645178124973" $x^{2} + \frac{7 x}{2} + {(\frac{1}{2} \cdot (\frac{7}{2}))}^{2} = \frac{15}{2} + {(\frac{1}{2} \cdot (\frac{7}{2}))}^{2}$

$x^{2} + \frac{7 x}{2} + {(\frac{7}{4})}^{2} = \frac{15}{2} + {(\frac{7}{4})}^{2}$

$x^{2} + \frac{7 x}{2} + \frac{49}{16} = \frac{15}{2} + \frac{49}{16}$

${(x + \frac{7}{4})}^{2} = \frac{169}{16}$

$\sqrt{{(x + \frac{7}{4})}^{2}} = \sqrt{\frac{169}{16}}$

$x + \frac{7}{4} = \pm \frac{13}{4}$

$x + \frac{7}{4} - \frac{7}{4} = - \frac{13}{4} - \frac{7}{4}, x + \frac{7}{4} - \frac{7}{4} = \frac{13}{4} - \frac{7}{4}$

$x = - \frac{20}{4}, x = \frac{6}{4}$

$x = - 5, x = \frac{3}{2}$

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