SAT Mathematics Practice Test 2

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