First slide

Theory of equations

Question

If $α$ and $β$ are the roots of the equation $x^{2} + a x + b = 0$ , and $α^{4}$ and $β^{4}$ are the roots of $x^{2} - p x + q = 0$ , the roots $x^{2} - 4 b x + 2 b^{2} - p = 0$ are always

Moderate

A

both non-real
B

both positive
C

both negative
D

positive and negative

Solution

We have,
$α + β = - a, α β = b, α^{4} + β^{4} = p$ and $α^{4} β^{4} = q .$
Now,
   $\begin{array}{l} α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2} \\ \Rightarrow α^{4} + β^{4} = {[(α + β)^{2} - 2 α β]}^{2} - 2 (α β)^{2} \\ \Rightarrow α^{4} + β^{4} = {(a^{2} - 2 b)}^{2} - 2 b^{2} [∵ α + β = - a, α β = b] \\ \Rightarrow α^{4} + β^{4} = {(a^{2})}^{2} - 4 a^{2} b + 2 b^{2} \\ \Rightarrow p = {(a^{2})}^{2} - 4 b (a^{2}) + 2 b^{2} [∵ α^{4} + β^{4} = p] \end{array}$
   $\Rightarrow {(a^{2})}^{2} - 4 b (a^{2}) + 2 b^{2} - p = 0$ …(i)
   $\Rightarrow a^{2}$ is a root of the equation $x^{2} - 4 b x + 2 b^{2} - p = 0$
Let $D$ be the discriminant of $x^{2} - 4 b x + 2 b^{2} - p = 0 .$ Then,
   $D = 16 b^{2} - 4 (2 b^{2} - p) = 16 b^{2} + 4 \{{(a^{2})}^{2} - 4 b (a^{2})\}$ [Using (i)]
$\Rightarrow D = 4 \{{(a^{2})}^{2} - 4 b a^{2} + 4 b^{2}\} = 4 {(a^{2} - 2 b)}^{2} \geq 0$
Hence, the roots of the given equation are real and are of opposite signs.

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