First slide

Theory of equations

Question

The equation whose roots are the square of the roots of the equation $2 x^{2} + 3 x + 1 = 0$

Easy

A

$4 x^{2} + 5 x + 1 = 0$
B

$4 x^{2} - x + 1 = 0$
C

$4 x^{2} - 5 x - 1 = 0$
D

$4 x^{2} - 5 x + 1 = 0$

Solution

Let the roots of equation
$2 x^{2} + 3 x + 1 = 0 is α and β$
Then , $α + β = - \frac{3}{2} and αβ = \frac{1}{2}$
$\begin{aligned} ∴ α^{2} + β^{2} = (α + β)^{2} - 2 αβ \\ = {(\frac{- 3}{2})}^{2} - 2 \times \frac{1}{2} = \frac{9}{4} - 1 = \frac{5}{4} \end{aligned}$
and $α^{2} β^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$
$∴$ Required equation is
$\begin{array}{l} x^{2} - (α^{2} + β^{2}) x + α^{2} β^{2} = 0 \\ \Rightarrow x^{2} - \frac{5}{4} x + \frac{1}{4} = 0 \\ \Rightarrow 4 x^{2} - 5 x + 1 = 0 \end{array}$

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