First slide

Theory of equations

Question

If $a, b, c \in R$ and the equations ${ax}^{2} + bx + c = 0$ and $x^{3} + 3 x^{2} + 3 x + 2 = 0$ have two roots in common, then

Moderate

A

$a = b \neq c$
B

$a = b = - c$
C

$a = b = c$
D

None of these

Solution

We have,
$\begin{array}{r} x^{3} + 3 x^{2} + 3 x + 2 = 0 \\ \Rightarrow (x + 1)^{3} + 1 = 0 \end{array}$
$\begin{array}{lc} \Rightarrow & (x + 1 + 1) [(x + 1)^{2} - (x + 1) + 1] = 0 \\ \Rightarrow & (x + 2) (x^{2} + x + 1) = 0 \\ \Rightarrow & x = - 2, \frac{- 1 \pm \sqrt{3} i}{2} \\ \Rightarrow & x = - 2, ω, ω^{2} \end{array}$
Since, $a, b, c \in R, {ax}^{2} + bx + c = 0$ cannot have one real and one imaginary root. Therefore, two common roots of ${ax}^{2} + bx + c = 0 and x^{3} + 3 x^{2} + 3 x + 2 = 0 are ω, ω^{2}$
$\begin{aligned} thus, - \frac{b}{a} & = ω + ω^{2} = - 1 \\ \Rightarrow a & = b \\ and \frac{c}{a} & = ω \cdot ω^{2} = 1 \Rightarrow c = a \\ \Rightarrow a & = b = c \end{aligned}$

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