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explanation please

Expert replies
by kaf » Thu Mar 19, 2009 12:11 am
Can some please explain to me why

sq root(x) = x-2

solving we get x=4 and x=1

thanks
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Source: — Problem Solving |

Re: explanation please

by Ian Stewart » Thu Mar 19, 2009 3:50 am
kaf wrote:Can some please explain to me why

sq root(x) = x-2

solving we get x=4 and x=1

thanks
If you square both sides, you have

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

We can solve this by factoring if we get zero on one side:

0 = x^2 - 5x + 4
0 = (x-4)(x-1)

If the product of those factors is zero, one of them must equal zero, so either x - 4 = 0 and x = 4, or x - 1 = 0 and x = 1.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by kaf » Thu Mar 19, 2009 8:20 am
Thanks a lot Ian

I saw this this DS question and i have been trying to figure it out.If you can please help with a little explanation i will be grateful
Are lines p (with slope m) and q (with slope n) perpendicular to each other?

1. m + 2 = n
2. m + n = 0

I don't know the answer to this question. I feel both the statements are required to answer the question (when combined the values of m & n are -1 & 1 respectively, which satisfies the perpendicular slope concept). Any contentions?
thanks
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by quocbao » Thu Mar 19, 2009 8:31 am
I think it should be x = 4

x = 1 is not possible, because square root x > 0, so x - 2 > 0, so x > 2
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by Ian Stewart » Thu Mar 19, 2009 8:40 am
Yes, quocbao is correct - you need to make sure that the solutions both make sense in the original equation. Since root(x) can't be negative, the solution x = 1 doesn't make sense here.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by gmat740 » Thu Mar 19, 2009 8:58 am
Are lines p (with slope m) and q (with slope n) perpendicular to each other?

1. m + 2 = n
2. m + n = 0

I don't know the answer to this question. I feel both the statements are required to answer the question (when combined the values of m & n are -1 & 1 respectively, which satisfies the perpendicular slope concept). Any contentions?
Well I am not an Expert like Ian but yes I do have a solution for this statement

Since its given that m and n are slopes and not the line
so m+2=n does not really gives a solution


However,
m+n=0
this gives m=-n
so slopes, m and n are equal and of opposite sign
Thus the condition of Perpendicularity is m*n=-1

But over here it's not given whether the value of m or n is 1

Untill its not given,then both answers are insuff(E)

Else B

Hope this helps
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by gmat740 » Thu Mar 19, 2009 12:17 pm
Image
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by Stuart@KaplanGMAT » Thu Mar 19, 2009 4:15 pm
gmat740 wrote: However,
m+n=0
this gives m=-n
so slopes, m and n are equal and of opposite sign
Thus the condition of Perpendicularity is m*n=-1
This question has been recently discussed here: https://www.beatthegmat.com/slope-t33092.html, but I wanted to address a small math error.

If m + n = 0

then:

m = -n

as you stated, BUT to get -1 on the right side we must DIVIDE both sides by n, leaving us with:

m/n = -1,

which is NOT the condition of perpendicularity (which is why statement (2) is insufficient by itself).
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by gmat740 » Thu Mar 19, 2009 5:12 pm
Thanks a lot Stuart, I got it.I mised out a point over here.
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by sanju09 » Tue Mar 24, 2009 3:36 am
Quote:
Are lines p (with slope m) and q (with slope n) perpendicular to each other?

1. m + 2 = n
2. m + n = 0

I don't know the answer to this question. I feel both the statements are required to answer the question (when combined the values of m & n are -1 & 1 respectively, which satisfies the perpendicular slope concept). Any contentions?
Statement 1 by itself is not sufficient to make us check whether or not m n = -1.

Same with statement 2.

Combining the two enables us to have m + 2 = -m or m = -1 and hence n = 1 and so m n = -1 is found to be yes, true! So how about C?
The mind is everything. What you think you become. –Lord Buddha



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