First slide

Theory of equations

Question

If the equation whose roots are the squares of the roots of the cubic $x^{3} - {ax}^{2} + bx - 1 = 0$ is identical with the given cubic equation, then

Moderate

A

$a = 0, b = 3$
B

$a = b = 0$
C

$a = b = 3$
D

a, b are roots $x^{2} + x + 2 = 0$

Solution

Given equation is $x^{3} - {ax}^{2} + bx - 1 = 0$ . If roots of the equation be $α, β, γ$ , then
$\begin{matrix} α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (αβ + βγ + γα) = a^{2} - 2 b \end{matrix}$
$\begin{array}{l} α^{2} β^{2} + β^{2} γ^{2} + γ^{2} α^{2} = (αβ + βγ + γα)^{2} - 2 αβγ (α + β + γ) = b^{2} - 2 a \\ α^{2} β^{2} γ^{2} = 1 \end{array}$
So, the equation whose roots are $α^{2}, β^{2}, γ^{2}$ is given by
$x^{3} - (a^{2} - 2 b) x^{2} + (b^{2} - 2 a) x - 1 = 0$
It is identical to
$\begin{array}{l} x^{3} - {ax}^{2} + bx - 1 = 0 \\ \Rightarrow a^{2} - 2 b = a and b^{2} - 2 a = b \end{array}$
Eliminating b, we get
$\frac{{(a^{2} - a)}^{2}}{4} - 2 a = \frac{a^{2} - a}{2}$
$\begin{array}{l} or a \{a (a - 1)^{2} - 8 - 2 (a - 1)\} = 0 \\ or a (a^{3} - 2 a^{2} - a - 6) = 0 \\ or a (a - 3) (a^{2} + a + 2) = 0 \\ \Rightarrow a = 0 or a = 3 or a^{2} + a + 2 = 0 \end{array}$
which gives b = 0 or b = 3 or b² + b + 2 = 0. So, a = b = 0 or a = b = 3 or a, b are roots of x² + x + 2 = 0.

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