Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

up vote 2 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.