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# If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0,

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Intern
Joined: 28 Jan 2013
Posts: 1
If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0, [#permalink]

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Updated on: 25 Sep 2013, 11:08
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If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0, what is the other factor of the equation?\

A. (x+2√3)
B. (x+4√3)
C. (x+6√3)
D. (x−2√3)
E. (x−6√3)

I tried to solve the equation by putting the value of x as (x−2√3) and got x= √48..and then I just got confused Can someone tell me the correct approach to the problem?

Thanks
[Reveal] Spoiler: OA

Originally posted by bzb on 25 Sep 2013, 10:59.
Last edited by Bunuel on 25 Sep 2013, 11:08, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 02 Sep 2009
Posts: 44573
Re: If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0, [#permalink]

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25 Sep 2013, 11:19
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bzb wrote:
If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0, what is the other factor of the equation?\

A. (x+2√3)
B. (x+4√3)
C. (x+6√3)
D. (x−2√3)
E. (x−6√3)

I tried to solve the equation by putting the value of x as (x−2√3) and got x= √48..and then I just got confused Can someone tell me the correct approach to the problem?

Thanks

Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$:

$$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$.

So, we know that one of the roots of $$x^2+(4\sqrt{3})x-36=0$$ is $$2\sqrt{3}$$.

According to the theorem above $$2\sqrt{3}*x_2=\frac{c}{a}=\frac{-36}{1}$$ --> $$x_2=-\frac{18}{\sqrt{3}}=-6\sqrt{3}$$.

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Re: If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0, [#permalink]

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23 Sep 2014, 22:33
Basic Formula:

(x+a) * (x+b) = x^2 + (a+b) * x + ab

Given equation:

$$x^2 + 4\sqrt{3} x - 36 = 0$$; one of the root = $$x - 2\sqrt{3}$$

Just observe & try to compare with the basic formula

1. Highlighted sign in the equation is +ve & middle sign of the root is -ve.

It means middle sign of the other root has to be +ve

Re-writing the equation as follow:

$$x^2 + (6\sqrt{3} - 2\sqrt{3}) x - (6\sqrt{3} * 2\sqrt{3}) = 0$$

$$x(x + 6\sqrt{3}) - 2\sqrt{3} (x + 6\sqrt{3}) = 0$$

$$(x + 6\sqrt{3}) (x - 2\sqrt{3}) = 0$$

$$Other root = x + 6\sqrt{3}$$

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Intern
Joined: 07 Feb 2016
Posts: 14
GMAT 1: 710 Q47 V40
Re: If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0, [#permalink]

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30 Aug 2016, 10:08
We don't really need Viete's theorem to solve this problem. We know that C=-36, so the second factor will be (x+k) for some number k. You can scratch D and E already. Now, the simple equation is -2√3 * k = -36. So K= -36/-2√3 --->6√3
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Re: If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0, [#permalink]

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11 Dec 2017, 03:33
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If (x−2√3) is one factor of the equation x^2+(4√3)·x−36=0,   [#permalink] 11 Dec 2017, 03:33
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