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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] New post 24 Oct 2012, 03:43
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B
C
D
E

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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, 04:02, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Quadratic Equation Root [#permalink] New post 24 Oct 2012, 04:08
Expert's post
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] New post 24 Oct 2012, 14:24
1
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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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Posts: 18721
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Re: Quadratic Equation Root [#permalink] New post 24 Oct 2012, 16:09
Expert's post
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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: Quadratic Equation Root   [#permalink] 24 Oct 2012, 16:09
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If q is one root of the equation x^2 + 18x + 11c = 0, where

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