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I do not know how to solve for $b$ in this expression:

$$\sqrt{\frac{1}{25} + b^2} = 1$$

My first guess was to multiply both sides by the left side, but then I do not see anything that looks more interesting to me. Actually my first guess was to square both sides which gives:

$$\frac{+}{-} \left( \frac{1}{25} + b^2 \right) = 1$$

What is some correct approach?

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The $\pm$ is unnecessary. It is true that if $\sqrt{x} = y$ then $x = y^2$. Besides, to discuss the square root of something, that something must be nonnegative. You seem to have confused this with the law $u^2 = v \implies \pm u = \sqrt v$. – Herng Yi Apr 6 '13 at 8:32
up vote 3 down vote accepted

$$\sqrt{\frac{1}{25} + b^2} = 1\iff \frac{1}{25} + b^2=1\iff b^2=\frac{24}{25}\iff b=\pm\frac{\sqrt{24}}{5}$$

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