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If q is one root of the equation x^2 + 18x + 11c = 0, where
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Updated on: 24 Oct 2012, 04:02
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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^2c^2 = A) 98 B) 72 C) 49 D) 0 E) It can't be determined from the information given
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Originally posted by LM on 24 Oct 2012, 03:43.
Last edited by Bunuel on 24 Oct 2012, 04:02, edited 1 time in total.
Renamed the topic and edited the question.




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Re: Quadratic Equation Root
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24 Oct 2012, 04:08




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Re: Quadratic Equation Root
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24 Oct 2012, 14:24
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 \(x_1*x_2=7*(11)=\frac{11c}{1}\) > \(c=7\). \(q^2c^2 =7^2(7)^2=0\). Answer: D. Hope it's clear. shouldn't q = 7 and c = 7?



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Re: Quadratic Equation Root
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24 Oct 2012, 16:09



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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where
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15 Sep 2014, 13:19
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^2c^2=(q+c)(qc), (q+c)=7+7=0, q^2c^2=0



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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where
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15 Sep 2014, 21:09
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^2c^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
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16 Sep 2014, 12:57
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^2c^2, because every number squared to an even power is a positive number, 11q=11c=>q^2=c^2, and therefore q^2c^2=0!



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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where
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22 Sep 2014, 01:44
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^2c^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
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05 May 2015, 12:51
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
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19 Feb 2017, 11:44
(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
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19 Feb 2017, 15:13
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^2c^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 (xq)(x+11)=x^2qx+11x11q=0 qx+11x=18x q=7 11q=11c c=7 (7)^27^2=0



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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where
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24 Feb 2017, 11:08
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^2c^2 =
A) 98 B) 72 C) 49 D) 0 E) It can't be determined from the information given Product of roots=c/a => \(q*11=\frac{11c}{1}\) =>\(c=q\) \(q^2c^2=q^2(q)^2=q^2q^2=0\)
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Re: If q is one root of the equation x^2 + 18x + 11c = 0, where
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