Question

How do I solve for 'k'?

Answer 1

`√(2k^2+17) -7=0`


`2k^2+17= \sqrt[ (1)/(2) ]{7}`


`2k^2+17=7^2`


`2k^2+17=49`


`2k^2=49-17`


`2k^2=32`


`k^2 = (32)/(2)`


`k^2=16`


`k= √(16)`


`k=4`