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OG QR 2nd Ed. #135

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OG QR 2nd Ed. #135

by tofubeans » Sat Apr 21, 2012 8:45 am
So in this problem I am asked to find the area of a square within a rectangle with some givens.

The equation I come out with is: x^2+12x-540=0 and all I would have to do is solve for x. However, looking at that seems a little time confusing if I don't already know what numbers to choose (30 and 18).

Is there a method to this that doesn't require so much time in looking for the correct numbers that multiply out to 540 and subtract to 12? Or would I just have to take it as it is and test out a bunch of numbers until I hit the mark?
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by aneesh.kg » Sat Apr 21, 2012 8:55 am
I discussed this exact problem in an earlier post.

The easiest way to factorise this equation is to try out the numbers but if you have practiced A LOT of quadratic equations, it is a 5 - 10 seconds job. Also, remember that, in GMAT you will NEVER have to solve a quadratic equation that is difficult to factorise. So, you will find your solution by trying out 3-4 values at maximum.

Suggestion: Grab a school textbook with a chapter on 'Quadratic Equations'. Solve all the problems on quadratic equation in it and you will become a champ in solving this. This is what I suggest to all my students who are not too comfortable with factorising Quadratic Equations.
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by Brent@GMATPrepNow » Sat Apr 21, 2012 10:46 am
tofubeans wrote:So in this problem I am asked to find the area of a square within a rectangle with some givens.

The equation I come out with is: x^2+12x-540=0 and all I would have to do is solve for x. However, looking at that seems a little time confusing if I don't already know what numbers to choose (30 and 18).

Is there a method to this that doesn't require so much time in looking for the correct numbers that multiply out to 540 and subtract to 12? Or would I just have to take it as it is and test out a bunch of numbers until I hit the mark?
So, we know we're essentially looking for 2 numbers such that the product is -540 and the sum is 12.
Or we might just say that the product is 540 and the difference is 12.

Okay, let's start with 10 and 54. (10)(54)=540, but the difference is 44 (no good).

Aside: If (10)(54)=540, we can double one value and halve the other value and the product will still be 540. Double 10 to get 20, and halve 54 to get 27.

So, we know that (20)(27)=540, but the difference is 7 (no good).

Let's go back to (10)(54)=540. Triple 10 to get 30, and find the third of 54 to get 18.
So, we know that (30)(18)=540, and the difference is 18 (bingo!).

Factor x^2+12x-540 to get (x+30)(x-18)=0 . . . etc.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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