GRE-Math-Questions-to-Know-by-Test-Day-2nd-Edition_pag241

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412. (D) x16 812 = 
 Solve for x 2 : x 81
16
2 = .
 Take the square root of both sides: x 9
4
= (because x > 0).
 Take the principal square root of both sides: x 3
2
= (D).
413. (B), (C), (D), (E) n 4= ± and m 6= ± , so the difference could be 10 (B)− , 2− (C), 
2 (D), or 10 (E).
414. (D) The square root symbol ( ) always indicates the principal (nonnegative) square 
root. Thus,
x ( 4)
7
4
7
2
2
= − = (D)
415. (A), (D) Substituting gives x x 22+ = , which is equivalent to x x 2 02 + − = . Factoring 
the left side of this equation yields x x( 2)( 1) 0+ − = , from which you have x = 2− (A) or 
x = 1 (D).
416. (B), (D) From y 42 = , y is 2 or 2− . Substitute each y value into the equation and solve 
for the corresponding value of x.
 Substituting y = 2 yields:
x x
x x
x
x
3.5 1.5(2) 1 0.5(2) 2.5
3.5 3 1 1 2.5
6 5
5
6
(B)
+ + = − −
+ + = − −
= −
= −
 
 Substituting y = 2− yields:
x x
x x
x
x
3.5 1.5( 2) 1 0.5( 2) 2.5
3.5 3 1 1 2.5
6 3
1
2
(D)
+ − + = − − −
− + = −
=
=
 
417. (E), (F) The graph of a parabola with vertex h k( , ) that opens upward has equa-
tion y x h k( )2= − + . The graph in the figure is a parabola with vertex ( 2,0)− , opening up. 
Thus, the equation is y x( 2)2= + . Select (E). Select (F) because y x x4 42= + + implies that 
y x( 2)2= + .
418. (C), (D), (E) x 1 2+