1) If x<0, then squareroot(-x|x|) is
a) -x b) -1 c) 1 d) x e)squareroot(x)
OA is A)
My approach is
since x<0 take x as -2 then sub in above equation, we get
squareroot(-(-2)|2|)= squareroot(4)= +- 2, since x<0 we get -2
we took x as -2 and we got answer as -2.
So answer is x, but in prep answer is given as A) ....why so????
2) What is the greatest possible area of a triangluar region with one vertex at the center of the circle of radius 1 and the other two vertices on the circle?
a) squareroot(3) / 4 b) 1/2 c) pie/4 d)1 e)squareroot(2)
OA is 1 /2 i.e B
Gmat prep problems
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(1) The example you presented is correct, up until:
"= squareroot(4)= +- 2"
On the GMAT, anything under a square root sign must yield a positive value. Negative Square roots are imaginary numbers and are outside the scope of the GMAT.
Therefore squareroot(4) simply yields a positive value of 2
Using your example, when -2 is inserted into the equation "squareroot(-x|x|)", the output is 2. Therefore the answer is (A) -X
"= squareroot(4)= +- 2"
On the GMAT, anything under a square root sign must yield a positive value. Negative Square roots are imaginary numbers and are outside the scope of the GMAT.
Therefore squareroot(4) simply yields a positive value of 2
Using your example, when -2 is inserted into the equation "squareroot(-x|x|)", the output is 2. Therefore the answer is (A) -X
- manpsingh87
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x<0; sqrt(-x|x|);Chaitanya_1986 wrote:1) If x<0, then squareroot(-x|x|) is
a) -x b) -1 c) 1 d) x e)squareroot(x)
as x<0; therefore |x|=-x;
sqrt(-x(-x)); sqrt(x^2); now this can be equal to either x or -x, but since x<0; therefore answer should be -x..!!!
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let's work through inequality condition and disallow x=0 (x is not equal to 0), √(-x*|x|)>0 as the negative root of numbers is not allowed in GMAT (according to GMAT conventions, it's allowed in MATH as a whole) -x*|x|>0 is possible only if x<0 because |x| is positive and -x multiplied by positive number will be less than 0. Therefore x must be negative, x<0. We agreed that √number cannot be negative in GMAT, hence √(-x*|x| is valid only with -x.Chaitanya_1986 wrote:1) If x<0, then squareroot(-x|x|) is
a) -x b) -1 c) 1 d) x e)squareroot(x)
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