The set of all real numbers x such that √(x^2) = -x consists of :
A zero only
B nonpositive real numbers only
C positive real numbers only
D all real numbers
E no real numbers
I thought the answer was E but the actual answer is B. For example if i put in √(-5)^2 in the cacl i get +5. So whats the explanation?
The square root of x^2 is the absolute value of x. To the left of zero, the absolute value of x = -x (graph the two functions and you’ll see this). Therefore, the answer is all negative numbers, or B.
lhw1998 is correct; the square root of x^2 is, by definition, the absolute value of x.
You were also just one step away from solving this problem by testing the answer choices, though. You plugged x = -5 into the expression and got +5 back. That’s consistent with (B), since you plugged a negative number in and got a positive number (-x) back.
You can therefore eliminate (A), since -5 is not zero. You can eliminate ©, since -5 is not a positive number. You can eliminate (E), since you know that the equation works when x = -5.
You can eliminate (D) by plugging in +5. You’ll get +5 back, which is not -x.
reformationtoday you said “since you plugged a negative number in and got a positive number (-x) back”. (-x) is a positive number? o_O
@lhw1998 small correction in your solution – the set of real numbers for which |x| = -x is the set of non-positive numbers, not negative numbers (they are almost equivalent, except that 0 is non-positive but not negative).
@ChemNerd123 If x is negative, then -x is positive.
Try putting parenthesis around x. √((x)^2) = -(x). If x = -2 then √((-2)^2) = -(-2) which is a true statement. The two negatives multiply to yield a positive number. In the OP, you said you plugged in -5, which becomes √((-5)^2) = -(-5) which is consistent with the equation.
oooooh okay. Thanks for the help everyone.