SAT Mathematics Practice Test 2

Since $\sqrt{2{\mathit{k}}^2+17}-\mathit{x}=0,$ and $\mathit{x}=7$, one can substitute 7 for x,which gives $\sqrt{2{\mathit{k}}^2+17}-7=0$ Adding 7 to each side of $\sqrt{2{\mathit{k}}^2+17}-7=0$ gives $\sqrt{2{\mathit{k}}^2+17}=7$,Squaring each side of $\sqrt{2{\mathit{k}}^2+17}=7$ will remove the square root symbol:${\left(\sqrt{2{\mathit{k}}^2+17}\right)}^2={\left(7\right)}^2$, or $2{\mathit{k}}^2+17=49$.Then subtracting 17 from each side of $2{\mathit{k}}^2+17=49$ give $2{\mathit{k}}^2=49-17=32$ and dividing each side of $2{\mathit{k}}^2=32$ by 2 gives ${\mathit{k}}^2=16$ gives $\mathit{k}=\pm 4$,and Since the problem states that $\mathit{k}>0$,it follows that$\mathit{k}=4$
Since the sides of an equation were squared while solving $\sqrt{2{\mathit{k}}^2+17}-7=0$,it is possible that an extraneous root was produced. However, substituting 4 for k in $\sqrt{2{\mathit{k}}^2+17}-7=0$ confirms that 4 is a solution for k:
Choices A,B and D are incorrect because substituting any of these values for $\mathit{k}$ in $\sqrt{2{\mathit{k}}^2+17}-7=0$ does not yield a true statement.