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So this is an equation from one of the solutions in my textbook that I am trying to understand as part of solving a cholesky-factorization problem:

$$\sqrt{18-(\frac{a}{\sqrt2})^2} = \sqrt{\frac{36-a^2}{\sqrt2}} $$

Which square root rule applies here? Feels like I am missing some basics...

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    $\begingroup$ The equation is not correct. The $\sqrt{2}$ on the right should be a $2$. $\endgroup$ – John Wayland Bales Mar 1 at 20:01
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That's because it isn't true. $$\sqrt{18-\left(\frac a{\sqrt2}\right)^2} = \sqrt{\frac{36-a^2}2}$$

Notice the $2$ in the right hand side as opposed to $\sqrt2$

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    $\begingroup$ Thank you, this drove me crazy! $\endgroup$ – Lucky Mar 1 at 20:18
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$$\sqrt{18-\left(\frac{a}{\sqrt2}\right)^2} = \sqrt{18-\frac{a^2}{2}} = \sqrt{\frac{36-a^2}{2}};$$

as pointed out by @John Wayland Bales in a comment to the question, $\sqrt2$ on the right side of the equation in the question should be $2$.

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It is $$18-\left(\frac{a}{\sqrt{2}}\right)^2=18-\frac{a^2}{2}=\frac{36-a^2}{2}$$

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  • $\begingroup$ I think you meant $36 \mathbf - a^2$ $\endgroup$ – J. W. Tanner Mar 1 at 20:13
  • $\begingroup$ Yes I meant this, thank you for your hint! $\endgroup$ – Dr. Sonnhard Graubner Mar 1 at 20:15

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