Absolute values

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Re: Absolute values

by billzhao » Mon Feb 16, 2009 1:43 am
Maratha1 wrote:Can anyone please explain the answer?

What is the value of |x|?

(1) |x2 + 16| – 5 = 27

(2) x2 = 8x – 16
From (1) |x^2+16|=32 => x^2=16 => x=4 or -4 =>|x|=4, suff

From (2) x^2-8*x+16=0 => (x-4)^2=0 =>x=4=>|x|=4, suff

Thus Answer is (D)
Yiliang

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Re: Absolute values

by Maratha1 » Mon Feb 16, 2009 3:36 am
billzhao wrote:
Maratha1 wrote:Can anyone please explain the answer?

What is the value of |x|?

(1) |x2 + 16| – 5 = 27

(2) x2 = 8x – 16
From (1) |x^2+16|=32 => x^2=16 => x=4 or -4 =>|x|=4, suff

From (2) x^2-8*x+16=0 => (x-4)^2=0 =>x=4=>|x|=4, suff

Thus Answer is (D)
Thanks dude!

I've one question though, why |x^2+16| this can not be +(x^2+16) and -(x^2+16)?
carpe diem!

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by billzhao » Mon Feb 16, 2009 3:57 am
I've one question though, why |x^2+16| this can not be +(x^2+16) and -(x^2+16)?

Because x^2>=0, |x^2+16| must be greater than 0 and must equal to x^2+16
Yiliang

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by AJ2009 » Mon Feb 16, 2009 4:29 pm
I go with D.

|x^2+16| is always a positive value so no need worry about signs.

therefore |x^2+16|=32 or x^2+16-32=0 so x=-4 or |x|=4

stmt II give on positive value of 4 so also sufficient.