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Re: (x+2)(x-2) [#permalink]
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Expert Reply
Carcass wrote:
Quantity A
Quantity B
\((x+2)(x-2)\)
\(\frac{-5(x^2-25)(x-1)}{(x+5)(x^2-6x+5)}\)


Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.



A=(\((x+2)(x-2)=x^2-4\)).. min value will be -4 as \(x^2\geq{0}\)
B=\(\frac{-5(x^2-25)(x-1)}{(x+5)(x^2-6x+5)}=\frac{-5(x-5)(x+5)(x-1)}{(x+5)(x^2-5x-x-5)}=\frac{-5(x+5)(x-1)(x-5)}{(x+5)(x-1)(x-5)}=-5\)

So A>B

A
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Re: (x+2)(x-2) [#permalink]
Answer: A
We just try to simplify them:

A= (x+2)(x-2) = x^2-4

B= -5 * (x^2 - 25)(x-1) / (x+5)(x^2-6x+5) =
-5 * (x-5)(x+5)(x-1) / (x+5)(x-5)(x-1) =
-5

A= x^2 -4 and B=-5, as x^2 is something positive, it can at least be zero, so x^2-4 is definitely more than -4, So A is bigger than B.
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Re: (x+2)(x-2) [#permalink]
quantity A: x^2-4
if x=0 then also value is -4

quantity 2: after solving the equation we will get -5

hence option A is the correct answer.
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Re: (x+2)(x-2) [#permalink]
1
QA -> x^2 -4
QB -> -5

Add +4 in QA and QB
QA -> x^2
QB -> -1

we know that square of any number is always greater than or equal to 0.

Therefore, QA > QB
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Re: (x+2)(x-2) [#permalink]
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