# How do you solve 1/(6b^2)+1/(6b)=1/(b^2 and check for extraneous solutions?

Nov 27, 2016

$b = 5$

Extraneous solution is $b = 0$

#### Explanation:

Write as:$\text{ } \frac{1}{6 {b}^{2}} + \frac{b}{6 {b}^{2}} = \frac{1}{b} ^ 2$

$\frac{1 + b}{6 {b}^{2}} = \frac{1}{b} ^ 2$

${b}^{2} \left(1 + b\right) = 6 {b}^{2}$

$1 + b = 6$

$b = 5$
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We are 'not allowed' to have 0 as a denominator. The proper name for this is that the expression is 'undefined'

So the extraneous solution is $b = 0$

$\frac{1}{6 {b}^{2}} + \frac{1}{6 b} = \frac{1}{b} ^ 2$

For $b = 0$ we have

$\frac{1}{0} + \frac{1}{0} = \frac{1}{0}$ which is undefined