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Updated on: 09 Jul 2013, 10:02
Originally posted by krishnasty on 20 Dec 2010, 08:20.
Last edited by Bunuel on 09 Jul 2013, 10:02, edited 1 time in total.
Last edited by Bunuel on 09 Jul 2013, 10:02, edited 1 time in total.
Edited the question.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (02:01) correct
18%
(02:10)
wrong
based on 449
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John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?
A. x^2 + 4x + 14 = 0
B. 2x^2 + 7x -24 = 0
C. x^2 -14x + 48 = 0
D. 3x^2 -17x + 52 = 0
E. 2x^2 + 4x + 14 = 0
A. x^2 + 4x + 14 = 0
B. 2x^2 + 7x -24 = 0
C. x^2 -14x + 48 = 0
D. 3x^2 -17x + 52 = 0
E. 2x^2 + 4x + 14 = 0
20 Dec 2010, 08:41
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krishnasty
Viete's formula for the roots \(x_1\) and \(x_2\) of equation \(ax^2+bx+c=0\): \(x_1+x_2=-\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).
John made a mistake while copying the constant term and got the roots as 5 and 9 so he copied coefficient of x and x^2 correctly: \(5+9=14=-\frac{b}{a}\);
Jane made a mistake in the coefficient of x and she got the roots as 12 and 4 so she copied the constant term and the coefficient of x^2 correctly: \(12*4=48=\frac{c}{a}\);
Only option C satisfies the above conditions \(x^2-14x+48=0\): \(-\frac{b}{a}=-\frac{-14}{1}=14\) and \(\frac{c}{a}=\frac{48}{1}=48\) (any of the condition would be sufficient).
Answer: C.
General Discussion
21 Dec 2010, 04:56
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Bunuel
I just solve it with (x-y)(x-y) aproach not sure if I was just lucky and landed on the right answer or if it does always work
So: John got values 9 and 5 for x; (x-5)(x-9)=0 --> x^2-14x+45=0 (his mistake 45)
Jane got values 12 and 4 for x; (x-12)(x-4)=0 --> x^2-16x+48=0 (her mistake 16x)
Eliminate mistakes and plug in the right values each picked: X^2-14x(from Jonhs equ)+48(from janes)=0
Guess that this might only work when X^2 does not have a multiplier...
Will be adding to my "Blackbook" Viete's formula for the roots.
