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# Quick way for this quadratic

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Director
Joined: 08 Jul 2004
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28 Jul 2004, 02:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Does anybody know how can i answer this quickly.

14. If R is a root of the equation X2 +3X â€“ 54, than which of the following equations have also the root R ?
X2 â€“ 12X +27.
X2 â€“ 6X â€“ 16.
X2 â€“ 10X â€“ 31.25.
X2 â€“ 15X + 54.
X2 + 10X + 16.

The original equation is X2 + 3X â€“ 54, it can be written as (X â€“ 6)(X + 9). The roots are 6 and (-9).
We are looking for an equation that has one of the same roots.
Answer D: X2 â€“ 15X +54 = (X â€“ 6)(X â€“ 9) â€¡ This equation has the root 6.
All the other answers have different roots than the original equation.
Intern
Joined: 27 Jul 2004
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28 Jul 2004, 08:25
I think your way is quick enough.

A. X2 – 12X +27=(x-3)(x-9)
B. X2 – 6X – 16=(x-8)(x+2)
D. X2 – 15X + 54=(x-6)(x-9)
E. X2 + 10X + 16=(x+2)(x+8)

You can quickly get those equations without calulating them. That is easy and quick.
CIO
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28 Jul 2004, 12:59
i would go right for D. It's also got a 54. It could be a trap, but trying it shows that it's not.
GMAT Club Legend
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28 Jul 2004, 22:39
I feel the best and fastest (not to mention surest ) way is to solve for the roots of the equation x^12 + 3x-54, which would come out ot be (x-6)(x+9). Then use backsolving from this step on. Using any one of the 2 roots, substitue back into the equation in the choice and if it adds up to 0, then that's the answer.
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