SATHEE: Theory of Equations 1 Question 31

Theory of Equations 1 Question 31

32. 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

(1989, 2M)

(a) two real roots

(b) two positive roots

(d) one positive and one negative root

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Solution:

Since, $α, β$ are the roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$ .

$\Rightarrow α + β = - p; α β = q; α^{4} + β^{4} = r$ and $α^{4} β^{4} = s$

Let roots of $x^{2} - 4 q x + (2 q^{2} - r) = 0$ be $α^{'}$ and $β^{'}$

Now, $α^{'} β^{'} = (2 q^{2} - r) = 2 (α β)^{2} - (α^{4} + β^{4})$

$= - (α^{4} + β^{4} - 2 α^{2} β^{2}) = - {(α^{2} - β^{2})}^{2} < 0$

$\Rightarrow$ Roots are real and of opposite sign.

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Theory of Equations 1 Question 31

32. 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

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32. 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