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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 6: Q. 435 (page 636)

In the following exercises, solve.
$16 p^{3} = 24 p^{2} - 9 p$

Short Answer

Expert verified

Answer is $p = 0, p = \frac{3}{4}$

Step by step solution

01

Step 1. Given

The equation is $16 p^{3} = 24 p^{2} - 9 p$

02

Step 2. Simplify the equation

$16 p^{3} = 24 p^{2} - 9 p 16 p^{3} - 24 p^{2} + 9 p = 0 p (16 p^{2} - 24 p + 9) = 0 p {(4 p - 3)}^{2} = 0$

So,

$p = 0$

and

$4 p - 3 = 0 p = \frac{3}{4}$

03

Step 3. Check both the values

Put $p = 0$ in the original equation,

$16 p^{3} = 24 p^{2} - 9 p 16 {(0)}^{3} = 24 {(0)}^{2} - 9 (0) 0 = 0$

True

Now put $p = \frac{3}{4}$ in the original equation,

$16 p^{3} = 24 p^{2} - 9 p 16 {(\frac{3}{4})}^{3} = 24 {(\frac{3}{4})}^{2} - 9 (\frac{3}{4}) 16 脳 \frac{27}{64} = 24 脳 \frac{9}{16} - \frac{27}{4} \frac{27}{4} = \frac{54}{4} - \frac{27}{4} \frac{27}{54} = \frac{27}{54}$

True

04

Step 4. Required answer

Answer is $p = 0, p = \frac{3}{4}$

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