If - be the roots of x2-px-q-0 and - - h- -h are the roots of x2-rx-s-0- then

Α + β= p, αβ=q α+β+2h= r, (α + h)(β+h)=s p+2h= r ⇒ h=p r/2 nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ........(i ...

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Standard XII

Mathematics

Algebra of Roots of Quadratic Equations

If α,β be the...

Question

If $α, β$ be the roots of $x^{2} + p x + q = 0$ and $α + h, β + h$ are the roots of $x^{2} + r x + s = 0$ , then

$p r^{2} = q s^{2}$

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$2 h = [\frac{p}{q} + \frac{r}{s}]$

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$p^{2} - 4 q = r^{2} - 4 s$

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$\frac{p}{r} = \frac{q}{s}$

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Solution

The correct option is D
$\frac{p}{r} = \frac{q}{s}$

$α + β = - p, α β = q α + β + 2 h = - r, (α + h) (β + h) = s$

$- p + 2 h = - r \Rightarrow h = \frac{p - r}{2}$ ........(i)

$N o w, α β + h (α + β) + h^{2} = s \Rightarrow q + h (- p) + h^{2} = s \Rightarrow q + (\frac{p - r}{2}) (- p) + {(\frac{p - r}{2})}^{2} = s \Rightarrow q + \frac{(- p^{2} + p r)}{2} + \frac{p^{2} + r^{2} - 2 p r}{4} = s$

$\Rightarrow 4 q - 2 p^{2} + 2 p r + p^{2} + r^{2} - 2 p r = 4 s \Rightarrow 4 q - p^{2} + r^{2} - 4 s = 0 \Rightarrow r^{2} - 4 s = p^{2} - 4 q$

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If $α, β$ be the roots of $x^{2}$ +px+q=0 and $α + h, β + h$ are the roots of $x^{2}$ +rx+s=0 , then

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If α,β be the roots of x2+px+q=0 and α+h, β+h are the roots of x2+rx+s=0, then

The correct option is D pr=qs α+β=−p, αβ=qα+β+2h=−r,(α+h)(β+h)=s −p+2h=−r⇒h=p−r2 ........(i) Now, αβ+h(α+β)+h2=s⇒q+h(−p)+h2=s⇒q+(p−r2)(−p)+(p−r2)2=s⇒q+(−p2+pr)2+p2+r2−2pr4=s ⇒4q−2p2+2pr+p2+r2−2pr=4s⇒4q−p2+r2−4s=0⇒r2−4s=p2−4q

If $α, β$ be the roots of $x^{2} + p x + q = 0$ and $α + h, β + h$ are the roots of $x^{2} + r x + s = 0$ , then