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23 Jun 2021, 05:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
66% (02:20) correct
34%
(02:11)
wrong
based on 175
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If the roots of the equation \(ax^2 + bx + c = 0\) are reciprocal of the roots of the equation \(px^2 + qx + r = 0\), then which of the following represents relation(s) between a, b, c, p, q and r?
(A) a = 1/p
(B) b = 1/q
(C) c = 1/r
(D) a = p, c = r and b = 1
(E) a = r, c = p and b = q
(A) a = 1/p
(B) b = 1/q
(C) c = 1/r
(D) a = p, c = r and b = 1
(E) a = r, c = p and b = q
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23 Jun 2021, 07:31
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Bunuel
We can work this problem backwards, let the two roots be \(r_1\) and \(r_2\), the reciprocal of them would be \(\frac{1}{r_1}\) and \(\frac{1}{r_2}\). Then the equations would be:
\((x - r_1)(x -r_2) = x^2 - (r_1 + r_2) + r_1*r_2\)
\((x - \frac{1}{r_1})(x - \frac{1}{r_2}) = x^2 -(\frac{1}{r_1} + \frac{1}{r_2})x + \frac{1}{r_1*r_2} = x^2 -(\frac{r_1 + r_2}{r_1*r_2})x + \frac{1}{r_1*r_2}\)
Now let's go back to the variables above, we can see the constants for both have a relation: \(r_1*r_2 = \frac{c}{a}\) and \(\frac{1}{r_1*r_2} = \frac{r}{p}\). Then \(\frac{c}{a} = \frac{p}{r}\) and \(\frac{c}{p} = \frac{a}{r}\).
For the middle term we have \(- (r_1 + r_2) = \frac{b}{a}\) and \(- \frac{r_1 + r_2}{r_1*r_2} = \frac{q}{p}\). The second equation we can plug in everything we learned earlier to get \(\frac{b}{a}*\frac{a}{c} = \frac{q}{p}\) then \(\frac{b}{c} = \frac{q}{p}\). We can write this as \(\frac{b}{q} = \frac{c}{p}\).
Thus finally we can concluded \(\frac{c}{p} = \frac{a}{r} = \frac{b}{q}\). With this the closest answer seems to be E. However E shoudln't be entirely correct since we can infinitely scale both equations, and we shouldn't be able to prove strong statements such as a = r.
23 Jun 2021, 07:49
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Bunuel
No answer choice is correct, which is easy to see, because we can freely multiply either quadratic equation by whatever we want without changing its roots. Here, our first quadratic could be
x^2 - 2x + 1 = 0
which has the single repeated root x = 1. The reciprocal of that root is still 1, so our second quadratic could be
3x^2 - 6x + 3 = 0
which is identical to the first quadratic, just multiplied by 3. Now it's easy to see none of the answer choices is right.
In a general quadratic, ax^2 + bx + c = 0, then c/a is the product of the roots. In the question asked, since the roots of the second quadratic are reciprocals of the roots of the first, the product of the roots of the first quadratic will be the reciprocal of the product of the roots of the second, or in other words, c/a = p/r. So answer C would be right here if we knew the quadratics looked like x^2 + bx + c and x^2 + qx + r, but as soon as we multiply the x^2 term by something, things get more complicated.
I think I've figured out what the question is getting at, but as it's written, it's wrong. If you also use the fact that the sum of the roots of ax^2 + bx + c is equal to -b/a, then if those roots are s and t, the sum of the roots of the other quadratic are 1/s + 1/t = (s + t)/st, and since s+t is the sum and st the product of the roots of the first quadratic, they equal -b/a and c/a respectively, so 1/s + 1/t = (-b/a)/(c/a) = -b/c. But we also know the sum of the roots of px^2 + qx + r must be -q/p. So -b/c = -q/p, and b/c = q/p.
So we know c/a = p/r and c/b = p/q, from which we see the ratio of c to b to a is identical to the ratio of p to q to r. That means if c = a, then b = q and a = r. It certainly does not mean that c= a, b = q, and a = r, however. All we can say is that c = ma, b = mq, and a = mr for some constant m.
So I'm guessing the "OA" is E here, but their math is wrong.
