amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation \(ax(cx+d)=-b(cx+d)\) is solved for x, which of the following is a possible ratio of the 2 solutions?
A. \(-\frac{ab}{cd}\)
B. \(-\frac{ac}{bd}\)
C. \(-\frac{ad}{bc}\)
D. \(\frac{ab}{cd}\)
E. \(\frac{ad}{bc}\)
Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!
Hi,
one method would be take advantage of SUM and PRODUCT of roots of a quadratic equation..\(ax(cx+d)=-b(cx+d)\) ..
\(acx^2+(ad+bc)x+bd = 0\)..
Let the roots be S and T
Sum of roots = \(S+T = -(\frac{ad+bc}{ac})\)..
Product of roots = \(S*T = \frac{bd}{ac}\)..
two methods..
1) equate from SUM..we know \(S+T = -(\frac{ad+bc}{ac})\)..
\(S+T = -(\frac{ad}{ac}+\frac{bc}{ac})\)
let \(S = -\frac{ad}{ac}\)and \(T = -\frac{bc}{ac}\)
so\(\frac{S}{T} = \frac{ad}{bc}\)
E
2)we are to find\(\frac{S}{T} or \frac{T}{S}\)..
divide SUM by PRODUCT--\(\frac{S+T}{ST} = -(\frac{ad+bc}{ac})/\frac{bd}{ac}\)..
\(\frac{S}{ST}+\frac{T}{ST} = -( \frac{ad}{bd} + \frac{bc}{bd})\) ..
\(\frac{1}{T} + \frac{1}{S} = -\frac{a}{b} - \frac{c}{d}\)..
so JUST for finding the ratio, we will equate two sides--
Here it doesn't matter what do you equate with, because we are just finding any of the ratio
\(\frac{1}{T}= -\frac{a}{b}\)and \(\frac{1}{S}= -\frac{c}{d}\)..
\(\frac{\frac{1}{T}}{\frac{1}{S}}\) = \(\frac{-\frac{a}{b}}{-\frac{c}{d}}\)..
\(\frac{S}{T} = \frac{ad}{bc}\)
E
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