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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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Updated on: 24 Oct 2012, 05:02
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45% (medium)

Question Stats:

72% (02:06) correct 28% (02:30) wrong based on 355 sessions

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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

Originally posted by LM on 24 Oct 2012, 04:43.
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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24 Oct 2012, 05:08
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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 $$x_1*x_2=(-7)*(-11)=\frac{11c}{1}$$ --> $$c=7$$.

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

Hope it's clear.
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24 Oct 2012, 15:24
1
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 $$x_1*x_2=7*(-11)=\frac{11c}{1}$$ --> $$c=-7$$.

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

Hope it's clear.

shouldn't q = -7 and c = 7?
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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 $$x_1*x_2=7*(-11)=\frac{11c}{1}$$ --> $$c=-7$$.

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

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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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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15 Sep 2014, 22:09
4
1
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.
$$(-7)^2 - 7^2 = 0$$

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

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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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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

$$c^2 - q^2 = 49 - 49 = 0$$

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

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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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19 Feb 2017, 12:44
1
(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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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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24 Feb 2017, 12:08
1
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]

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14 Nov 2018, 08:14
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where   [#permalink] 14 Nov 2018, 08:14
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