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

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New post 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 :cry: Can someone tell me the correct approach to the problem?

Thanks

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

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New post 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 :cry: 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 \(2\sqrt{3}*x_2=\frac{c}{a}=\frac{-36}{1}\) --> \(x_2=-\frac{18}{\sqrt{3}}=-6\sqrt{3}\).

Answer: C.
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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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New post 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:

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

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

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

\(Other root = x + 6\sqrt{3}\)

Answer = C
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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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New post 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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New post 11 Dec 2017, 03:33
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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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