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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 11: Q 6.98 (page 1069)

Solve $16 y^{2} = 32 y^{3} + 2 y$

Short Answer

Expert verified

The value is $y = \frac{1}{4}$

Step by step solution

01

Given information

Given the equation $16 y^{2} = 32 y^{3} + 2 y$

02

Bring all terms to one side and solve

Here, we have

$16 y^{2} = 32 y^{3} + 2 y 32 y^{3} + 2 y - 16 y^{2} = 0 32 y^{3} - 16 y^{2} + 2 y = 0 2 y (16 y^{2} - 8 y + 1) = 0$

03

Use zero factor property and calculate the value

Further solving,

$16 y^{2} - 8 y + 1 = 0 16 y^{2} - 4 y - 4 y + 1 = 0 4 y (4 y - 1) - 1 (4 y - 1) = 0 (4 y - 1) (4 y - 1) = 0 {(4 y - 1)}^{2} = 0 鈬� y = \frac{1}{4}$

04

Substitute and check the value

Substituting and checking, we get

$16 {(\frac{1}{4})}^{2} = 32 {(\frac{1}{4})}^{3} + 2 (\frac{1}{4}) 1 = \frac{1}{2} + \frac{1}{2} 1 = 1$

So the value is correct

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