If - and - are the roots of x2-px-q-0 and - 4- - 4 are the roots of x2-rx-s-0- then the equation x2-4qx-2q2-r-0- has always two real roots-

Form 1st we have nbsp; α ^2+β ^2=p^2 2q, and from 2nd nbsp; α ^4+β ^4=r ⇒ nbsp; ( α ^2+β ^2 )^2 2α ^2β ^2=r nbsp;⇒ nbsp; ( p^22q )^2=r ⋯ ( 1 ...

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Standard X

Mathematics

Roots of Quadratic Equation

If α and ...

Question

If $α$ and $β$ are the roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0,$ then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0,$ has always two real roots.

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Solution

Form 1st we have $α^{2} + β^{2} = p^{2} - 2 q,$ and from 2nd
$α^{4} + β^{4} = r \Rightarrow {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2} = r \Rightarrow {(p^{2} 2 q)}^{2} = r \dots (1)$
$x^{2} - 4 q x + (2 q^{2} - r) = 0$ ...(2)
Agian $Δ$ for the above equation is
$16 q^{2} - 4 (2 q^{2} - r) = (8 q^{2} + 4 r)$ $= 8 q^{2} + 4 {(p^{2} - 2 q^{2}) - 2 q^{2}}, b y (1)$ $= 4 {(p^{2} - 2 q)}^{2}$
which is + ive . Hence the roots of (2) are real roots.

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If α and β are the roots of x2+px+q=0 and α4,β4 are the roots of x2−rx+s=0, then the equation x2−4qx+2q2−r=0, has always two real roots.

Form 1st we have α2+β2=p2−2q, and from 2nd α4+β4=r⇒(α2+β2)2−2α2β2=r⇒(p22q)2=r⋯(1) x2−4qx+(2q2−r)=0 ...(2)Agian Δ for the above equation is 16q2−4(2q2−r)=(8q2+4r) =8q2+4{(p2−2q2)−2q2},by(1) =4(p2−2q)2 which is + ive . Hence the roots of (2) are real roots.

If $α$ and $β$ are the roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0,$ then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0,$ has always two real roots.