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Learn Pythagoras Theorem

From a handpicked tutor in live 1-to-1 classes

Sure, here’s how to solve for k:

    \[\sqrt{2k^2 + 5} = 8\]

To isolate k, we can square both sides of the equation:

    \[(\sqrt{2k^2 + 5})^2 = 8^2\]

Simplifying the left-hand side:

    \[2k^2 + 5 = 64\]

Subtracting 5 from both sides:

    \[2k^2 = 59\]

Dividing both sides by 2:

    \[k^2 = \frac{59}{2}\]

Taking the square root of both sides:

    \[k = \pm \sqrt{\frac{59}{2}}\]

So the solutions for k are:

    \[k = \sqrt{\frac{59}{2}} \approx 4.07\]

or

    \[k = -\sqrt{\frac{59}{2}} \approx -4.07\]

Note that we took the positive and negative square root because the original equation was an equation of a square root.