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05 Jun 2020, 08:55
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:32) correct
41%
(02:44)
wrong
based on 209
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If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?
A. -7
B. −5/2
C. 5/2
D. 7
E. 21
A. -7
B. −5/2
C. 5/2
D. 7
E. 21
05 Jun 2020, 09:06
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If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?
A. -7 --> correct: let's say p & q are the roots of the equation of \(x^2 + 28x – 588 = 0\) => (x-p)(x-q)=0 => x^2-(p+q)x+pq = 0, so p+q=-28, And let's say s & t are the roots of the equation of \(x^2 + bx + 98 = 0\) => so s*t = 98 and s = (p+q)/2=-28/2=-14, so -14t=98=>t = -7, so the other root of the equation \(x^2 + bx + 98 = 0\)is -7
B. −5/2
C. 5/2
D. 7
E. 21
A. -7 --> correct: let's say p & q are the roots of the equation of \(x^2 + 28x – 588 = 0\) => (x-p)(x-q)=0 => x^2-(p+q)x+pq = 0, so p+q=-28, And let's say s & t are the roots of the equation of \(x^2 + bx + 98 = 0\) => so s*t = 98 and s = (p+q)/2=-28/2=-14, so -14t=98=>t = -7, so the other root of the equation \(x^2 + bx + 98 = 0\)is -7
B. −5/2
C. 5/2
D. 7
E. 21
05 Jun 2020, 09:14
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Bunuel
Asked: If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?
Sum of roots of the equation (x^2 + 28x – 588 = 0 ) = -28
Average of the roots of the equation (x^2 + 28x – 588 = 0 ) = -28/2 = -14
One root of the equation (x^2 + bx + 98 = 0) = - 14
Let the other root of the equation (x^2 + bx + 98 = 0) be x
-14x = 98
x = - 7
IMO A
