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If 'k' be the ratio of the roots of the equation x^{2} - px +q

Question

If $^{'} k^{'}$ be the ratio of the roots of the equation $x^{2} - p x + q = 0$ , the value of $\frac{k}{1 + k ^{2}}$ is

\frac{q ^{2} - 2 p}{p}

\frac{q}{p ^{2} - 2 q}

\frac{p}{q ^{2} - p}

\frac{p}{p ^{2} - 2 q}

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^{'} k^{'}

be the ratio of the roots of the equation

x^{2} - p x + q = 0

, the value of

\frac{k}{1 + k ^{2}}

10 mins ago

Discuss this question LIVE

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Text solutionVerified

We know that equation $a x^{2} + b x + c = 0$

Then sum of roots $= \frac{- b}{a}$

Let the roots of the equation be a and b.
Then, a + b = p
ab = q

\frac{a}{b} = k

Now,

\frac{k}{1 + k ^{2}} = \frac{\frac{a}{b}}{1 + ( \frac{a}{b} ) ^{2}}

\frac{ab}{a ^{2} + b ^{2}}

\frac{ab}{( a + b ) ^{2} - 2 ab}

\frac{q}{p ^{2} - 2 q}

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Question Text	If $^{'} k^{'}$ be the ratio of the roots of the equation $x^{2} - p x + q = 0$ , the value of $\frac{k}{1 + k ^{2}}$ is
Answer Type	Text solution:1
Upvotes	151