If the ratio of the roots of the equation $ax^{2}+bx+c=0$ is $m : n$, then
- $ \displaystyle \frac{m}{n}+\frac{n}{m}=\frac{\ b ^2 }{ac}$
- $\sqrt{\left ( \displaystyle \frac{m}{n} \right )}+\sqrt{\left ( \displaystyle \frac{n}{m} \right )}=\displaystyle \frac{\left | b \right |}{\sqrt{c a }}$
- $\sqrt{\left ( \displaystyle \frac{m}{n} \right )}+\sqrt{\left (\displaystyle \frac{n}{m} \right )}=\displaystyle \frac{\ b ^2 }{\ c a}$
- $\sqrt{\left ( \displaystyle \frac{m}{n} \right )}+\sqrt{\left ( \displaystyle \frac{n}{m} \right )}=\displaystyle \frac{\ a ^2 }{\ b ^2}$
Reveal answer
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B
Correct answer
Explanation
Let roots be mr and nr. Sum = (m+n)r = -b/a, Product = mnr^2 = c/a. Squaring the sum and dividing by the product eliminates r, leading to the identity involving square roots of m/n and n/m.
AI explanation
Let the roots be mk and nk for some constant k. By Vieta's formulas, the sum of the roots is mk + nk = -b/a and the product is mk * nk = c/a. We can express k^2 = c/(mn a^2), which means k = sqrt(c/a) / sqrt(mn) because mn is positive. Substituting this into the ratio sum, sqrt(m/n) + sqrt(n/m) equals (m+n)/sqrt(mn). Since (mk+nk)^2 = b^2/a^2, we have (m+n)^2 * c/(mn a^2) = b^2/a^2, so (m+n)/sqrt(mn) = |b|/sqrt(ca). Therefore, the value of sqrt(m/n) + sqrt(n/m) is |b|/sqrt(ca).