Question

How do I solve for 'k'?

Answer 1

$\sqrt{2k^2+17} -7=0$ 

$2k^2+17= \sqrt[ \frac{1}{2} ]{7}$ 

$2k^2+17=7^2$ 

$2k^2+17=49$ 

$2k^2=49-17$ 

$2k^2=32$ 

$k^2 = \dfrac{32}{2}$ 

$k^2=16$ 

$k= \sqrt{16}$ 

$k=4$