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If q is one root of the equation x^2 + 18x + 11c = 0, where

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If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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If q is one root of the equation x^2 + 18x + 11c = 0, where –11 is the other root and c is a constant, then q^2-c^2 =

A) 98
B) 72
C) 49
D) 0
E) It can't be determined from the information given
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Oct 2012, 05:02, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Quadratic Equation Root [#permalink]

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LM wrote:
If q is one root of the equation \(x^2 + 18x + 11c = 0\), where –11 is the other root and c is a constant, then \((q^2)-(c^2) =\)

A) 98
B) 72
C) 49
D) 0
E) It can't be determined from the information given


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 \(x_1*x_2=\frac{c}{a}\).


Thus according to the above \(x_1+x_2=q+(-11)=\frac{-18}{1}\) --> \(q=-7\) AND \(x_1*x_2=(-7)*(-11)=\frac{11c}{1}\) --> \(c=7\).

\(q^2-c^2 =(-7)^2-7^2=0\).

Answer: D.

Hope it's clear.
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Re: Quadratic Equation Root [#permalink]

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Bunuel wrote:
LM wrote:
If q is one root of the equation \(x^2 + 18x + 11c = 0\), where –11 is the other root and c is a constant, then \((q^2)-(c^2) =\)

A) 98
B) 72
C) 49
D) 0
E) It can't be determined from the information given


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 \(x_1*x_2=\frac{c}{a}\).


Thus according to the above \(x_1+x_2=q+(-11)=\frac{-18}{1}\) --> \(q=7\) AND \(x_1*x_2=7*(-11)=\frac{11c}{1}\) --> \(c=-7\).

\(q^2-c^2 =7^2-(-7)^2=0\).

Answer: D.

Hope it's clear.



shouldn't q = -7 and c = 7?
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Re: Quadratic Equation Root [#permalink]

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New post 24 Oct 2012, 17:09
watwazdaquestion wrote:
Bunuel wrote:
LM wrote:
If q is one root of the equation \(x^2 + 18x + 11c = 0\), where –11 is the other root and c is a constant, then \((q^2)-(c^2) =\)

A) 98
B) 72
C) 49
D) 0
E) It can't be determined from the information given


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 \(x_1*x_2=\frac{c}{a}\).


Thus according to the above \(x_1+x_2=q+(-11)=\frac{-18}{1}\) --> \(q=7\) AND \(x_1*x_2=7*(-11)=\frac{11c}{1}\) --> \(c=-7\).

\(q^2-c^2 =7^2-(-7)^2=0\).

Answer: D.

Hope it's clear.



shouldn't q = -7 and c = 7?


Sure. Typo edited. Thank you. +1.
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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New post 15 Sep 2014, 14:19
The way I approached this question is as follows:

1) If one root is -11 then using the rule of factoring: (x+11)(x+7)=0, q=-7, c=7
2) Since q^2-c^2=(q+c)(q-c), (q+c)=-7+7=0, q^2-c^2=0
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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LM wrote:
If q is one root of the equation x^2 + 18x + 11c = 0, where –11 is the other root and c is a constant, then q^2-c^2 =

A) 98
B) 72
C) 49
D) 0
E) It can't be determined from the information given


Another method to solve it is - plug the root that is available to get the value of c.

\((-11)^2 + 18*(-11) + 11c = 0\)
\(c = 7\)

So the equation becomes \(x^2 + 18x + 77 = 0\) which gives x = -7 or -11.
So q must be -7.
\((-7)^2 - 7^2 = 0\)

Answer (D)
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If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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New post 16 Sep 2014, 13:57
one way to solve it really fast is to factor it:
x=-11 thus we get (x+11)
in order to get x^2+18x+11c=0, q must be also a negative, otherwise we cannot get to this equation, therefore we get x+q

now we can factor (x+q)(x+11)=0
q+11=18
q*11=11c

since we need q^2-c^2, because every number squared to an even power is a positive number, 11q=11c=>q^2=c^2, and therefore q^2-c^2=0!
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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New post 22 Sep 2014, 02:44
LM wrote:
If q is one root of the equation x^2 + 18x + 11c = 0, where –11 is the other root and c is a constant, then q^2-c^2 =

A) 98
B) 72
C) 49
D) 0
E) It can't be determined from the information given


\(x^2 + 18x + 11c = 0\)

\(x^2 + (11 + 7)x + (11 * c) = 0\)

c = 7

Given one root = -11, so other root = -7 = q

\(c^2 - q^2 = 49 - 49 = 0\)

Answer = D
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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New post 05 May 2015, 13:51
We have roots q and -11, so using vieta's theorem X1*X2 = C --> q*(-11) = 11c -> q = -c then q^2 - c^2 = 0
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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New post 19 Feb 2017, 12:44
(x+11)(x+q) = 0

11+q = 18
q= 7
11*q = 77
11*c = 77
c=7
q^2 -c^2 =(-7)^2 - 7^2 = 0
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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New post 19 Feb 2017, 16:13
LM wrote:
If q is one root of the equation x^2 + 18x + 11c = 0, where –11 is the other root and c is a constant, then q^2-c^2 =

A) 98
B) 72
C) 49
D) 0
E) It can't be determined from the information given


if q and -11 are roots,
then (x-q)(x+11)=x^2-qx+11x-11q=0
-qx+11x=18x
q=-7
-11q=11c
c=7
(-7)^2-7^2=0
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where [#permalink]

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New post 24 Feb 2017, 12:08
LM wrote:
If q is one root of the equation x^2 + 18x + 11c = 0, where –11 is the other root and c is a constant, then q^2-c^2 =

A) 98
B) 72
C) 49
D) 0
E) It can't be determined from the information given


Product of roots=c/a
=> \(q*-11=\frac{11c}{1}\)
=>\(c=-q\)

\(q^2-c^2=q^2-(-q)^2=q^2-q^2=0\)
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where   [#permalink] 24 Feb 2017, 12:08
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