First slide

Theory of equations

Question

Let $α and α^{2}$ be the roots of $x^{2} + x + 1 = 0$ then the equation whose roots are $α^{31} and α^{62}$ , is

Easy

A

$x^{2} - x + 1 = 0$
B

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

$x^{2} + x + 1 = 0$
D

$x^{60} + x^{30} + 1 = 0$

Solution

$Since, α, α^{2} be the roots of the equation x^{2} + x + 1 = 0$
$\begin{array}{lrl} ∴ α + α^{2} = - 1 \\ and α^{3} = 1 ...(i) \\ Now, α^{31} + α^{62} = α^{31} (1 + α^{31}) \\ \Rightarrow α^{31} + α^{62} = α^{30} \cdot α (1 + α^{30} \cdot α) \\ \Rightarrow α^{31} + α^{62} = {(α^{3})}^{10} \cdot α \{1 + {(α^{3})}^{10} \cdot α\} \\ \Rightarrow α^{31} + α^{62} = α (1 + α) [from Eq. (ii)] \\ \Rightarrow α^{31} + α^{62} = - 1 [from Eq. (i)] \\ Again, α^{31} \cdot α^{62} = α^{93} \\ \Rightarrow α^{31} \cdot α^{62} = {[α^{3}]}^{31} = 1 \end{array}$
$∴$ Required equation is,
$\begin{array}{l} x^{2} - (α^{31} + α^{62}) x + α^{31} \cdot α^{62} = 0 \\ \Rightarrow x^{2} + x + 1 = 0 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App