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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 9: Q. 107 E (page 885)

In the following exercises, solve by completing the square.
$2 p^{2} + 7 p = 14$

Short Answer

Expert verified

The solution is $p = - \frac{\sqrt{161}}{4} - \frac{7}{4}, p = \frac{\sqrt{161}}{4} - \frac{7}{4}$

Step by step solution

01

Step 1. Given information

Equation: $2 p^{2} + 7 p = 14$

02

Step 2. Calculation

We have,

$2 p^{2} + 7 p = 14$

$\frac{2 p^{2} + 7 p}{2} = \frac{14}{2}$

$p^{2} + \frac{7 p}{2} + {(\frac{1}{2} 路 (\frac{7}{2}))}^{2} = 7 + {(\frac{1}{2} 路 (\frac{7}{2}))}^{2}$

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

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

${(p + \frac{7}{4})}^{2} = \frac{161}{16}$

$\sqrt{{(p + \frac{7}{4})}^{2}} = \sqrt{\frac{161}{16}}$

$p + \frac{7}{4} = 卤 \frac{\sqrt{161}}{4}$

$p + \frac{7}{4} - \frac{7}{4} = - \frac{\sqrt{161}}{4} - \frac{7}{4}, p + \frac{7}{4} - \frac{7}{4} = \frac{\sqrt{161}}{4} - \frac{7}{4}$

localid="1645180152491" $p = - \frac{\sqrt{161}}{4} - \frac{7}{4}, p = \frac{\sqrt{161}}{4} - \frac{7}{4}$

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