Is x^2 5^2 ? (1) |x + 5| = 3|x - 5| (2) |x| 3

Question

Is x^2 > 5^2 ?

(1) |x + 5| = 3|x - 5|

(2) |x| > 3

Bunuel · Accepted Answer

Is x^2 > 5^2 ?

(1) |x+5| = 3|x - 5|

If x+5 and x-5 have the same sign, then we'd have \(x+5=3(x-5)\) --> \(x=10\). This value satisfies |x+5| = 3|x - 5|. So. \(x=10\) is a valid solution. This value gives an YES answer to the question.

If x+5 and x-5 have different signs, then we'd have \(-(x+5)=3(x-5)\) --> \(x=2.5\). This value satisfies |x+5| = 3|x - 5|. So. \(x=2.5\) is a valid solution. This value gives a NO answer to the question.

Not sufficient.

(2) |x| > 3. Clearly insufficient, consider x = 4 and x = 10.

(1)+(2) Since from (2) |x| > 3, then from (1) we are left with only x = 10. Therefore x^2 > 25. Sufficient.

Answer: C.

P.S. Please name the topics properly and provides the OA's (rules 3 and 7 here: rules-for-posting-please-read-this-before-posting-133935.html). Thank you.

madn800 · Answer

Then, how did you conclude that they have different signs?? Please explain it to me because I didn't understand that part.

Bunuel · Answer

|x+5| and 3|x - 5| can be expanded only in two ways, with the same sign and with different signs. Here is more conventional approach for you: (1) |x+5| = 3|x - 5|. Critical points are -5 and 5. If x\leq{-5} , then x+5\leq{0} , and x-5<0 , hence |x+5|=-(x+5) and x - 5=-(x-5) . Thus for this range the equation becomes: -(x+5)=-3(x-5) --> x=10 . Discard because this solution is out of the range. If -50 , and x-5<0 , hence |x+5|=x+5 and x - 5=-(x-5) . Thus for this range the equation becomes: x+5=-3(x-5) --> x=2.5 . Valid solution because it falls into the range. If x\geq{5} , then x+5>0 , and x-5\geq{0} , hence |x+5|=x+5 and x - 5=x-5 . Thus for this range the equation becomes: x+5=3(x-5) --> x=10 . Valid solution because it falls into the range. So, we have two solutions: x=2.5 and x=10 . Hope it helps.

MonishBhawale · Answer

another way can be
|x+5|=3|x-5|
square both
x2+10x +25 = 9x2-90x+225
2x2+25x+50=0
x=2.5orx=10

Is x^2 > 5^2 ? (1) |x + 5| = 3|x - 5| (2) |x| > 3

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