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Intermediate Algebra

OPENSTAX

$Math Studyset Vaia Explanations$ Math

OER 2017 Edition · 1346 pages

ISBN: 9780998625720

Chapter 11: Q. 6.91 (page 1069)

Solve $25 p^{2} = 49$ $25 p^{2} = 49$

Short Answer

Expert verified

The value is $p = \pm \frac{7}{5}$

Step by step solution

01

Given information

Given the equation, $25 p^{2} = 49$

02

Writing in standard form

Writing in standard form, we get

$25 p^{2} - 49 = 0 (5 p - 7) (5 p + 7) = 0 p = \frac{7}{5}, - \frac{7}{5}$

03

Checking the values

Checking the values, we get

$25 {(\frac{7}{5})}^{2} = 49 25 \times \frac{49}{25} = 49 49 = 49$

04

Checking another value

Checking the value, we get

$25 {(\frac{- 7}{5})}^{2} = 49 25 \times \frac{49}{25} = 49 49 = 49$

This is true

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Most popular questions from this chapter

In the following exercises, simplify each expression.

${(y^{4})}^{3} . {(y^{5})}^{2}$

In the following exercise, factor completely using the perfect square trinomials pattern.

$5 k^{3} - 70 k^{2} + 245 k$

In the following exercises, multiply. $(\sqrt{3} + \sqrt{10}) (\sqrt{3} + 2 \sqrt{10})$

Use the Quotient Property to simplify square roots.

(a) $\sqrt{\frac{72 x^{5}}{y^{6}}}$

(b) $\sqrt[5]{\frac{96 r^{11}}{s^{5}}}$

(c) $\sqrt[6]{\frac{128 u^{7}}{v^{12}}}$

In the following exercise, factor using substitution.

${(x - 3)}^{2} - 5 (x - 3) - 36$

See all solutions

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