Since
√2k2+17−x=0, and
x=7, one can substitute 7 for x,which gives
√2k2+17−7=0 Adding 7 to each side of
√2k2+17−7=0 gives
√2k2+17=7,Squaring each side of
√2k2+17=7 will remove the square root symbol:
(√2k2+17)2=(7)2, or
2k2+17=49.Then subtracting 17 from each side of
2k2+17=49 give
2k2=49−17=32 and dividing each side of
2k2=32 by 2 gives
k2=16 gives
k=±4,and Since the problem states that
k>0,it follows that
k=4Since the sides of an equation were squared while solving
√2k2+17−7=0,it is possible that an extraneous root was produced. However, substituting 4 for k in
√2k2+17−7=0 confirms that 4 is a solution for k:
√2(4)2+17−7=√32+17−7=√49−7=7−7=0
Choices A,B and D are incorrect because substituting any of these values for
k in
√2k2+17−7=0 does not yield a true statement.