If - - - are the roots of a x 2 - b x - c - 0 and - - h - - - h are the roots of p x 2 - q x - r - 0 - then h -

The correct option is D 1 2 ( b a q p ) Consider the given equation. ax^2+bx+c=0 #160; #160; #160; #160; #160; #160; #160 ...

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Standard XII

Mathematics

Location of Roots

If α , β ar...

Question

If $α, β$ are the roots of $a x^{2} + b x + c = 0$ and $α + h . β + h$ are the roots of $p x^{2} + q x + r = 0,$ then $h =$

\frac{b}{a} - \frac{q}{p}

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\frac{1}{2} (\frac{b}{a} - \frac{q}{p})

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- \frac{1}{2} (\frac{a}{b} - \frac{p}{q})

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- \frac{1}{2} (\frac{a}{b} + \frac{p}{q})

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Solution

The correct option is D $\frac{1}{2} (\frac{b}{a} - \frac{q}{p})$
Consider the given equation.
$a x^{2} + b x + c = 0$
$α + β = \frac{- b}{a}$ $. . . . . . (1)$

Since,
$p x^{2} + q x + r = 0$
$α + h + β + h = \frac{- q}{p}$
$α + β + 2 h = \frac{- q}{p}$

From equation $(1)$ , we get
$\frac{- b}{a} + 2 h = \frac{- q}{p}$
$2 h = \frac{b}{a} - \frac{q}{p}$
$h = \frac{1}{2} (\frac{b}{a} - \frac{q}{p})$

Hence, this is the answer.

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If α,β are the roots of ax2+bx+c=0 and α+h.β+h are the roots of px2+qx+r=0, then h=

The correct option is D 12(ba−qp)Consider the given equation.ax2+bx+c=0 α+β=−ba ......(1)Since,px2+qx+r=0α+h+β+h=−qpα+β+2h=−qpFrom equation (1), we get−ba+2h=−qp2h=ba−qph=12(ba−qp)Hence, this is the answer.

If $α, β$ are the roots of $a x^{2} + b x + c = 0$ and $α + h . β + h$ are the roots of $p x^{2} + q x + r = 0,$ then $h =$