Question

If $x^2+\lambda x+1=0,\lambda \in (-2,2)$ and $4x^3+3x+2x=0$ have common root then $c+\lambda$ can be

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $0$
• D. $\frac{3}{2}$

Answer 1

Answer: (A), (B)

The correct options are
A $-\frac{1}{2}$
B $\frac{1}{2}$

Given that,

$x^2+\lambda x+1=0$

On comparing that,

$a{x}^2+bx+c=0$

Then,

$a=1,b=\lambda ,c=1$

Now,

$D=b^2-4ac$

$D={\lambda }^2-4$

But given that,

$\lambda \in (-2,2)$

$\Rightarrow -2<\lambda <2$

$\Rightarrow 0<{\lambda }^2<4$

$\Rightarrow -4<{\lambda }^2-4<0$

Now,

$x^2+\lambda x+1=0$ is a factor of $4x^3+3x+2c=0$

So,

$4x^3+3x+2c=4x(x^2+\lambda x+1)-4\lambda (x^2+\lambda x+1)$

$4x^3+3x+2c=(4x-4\lambda )(x^2+\lambda x+1)$

On comparing coefficients of $x^1$

Now,

$3=4-4{\lambda }^2$

$\Rightarrow 4{\lambda }^2=1$

$\Rightarrow {\lambda }^2=\frac{1}{4}$

$\Rightarrow \lambda =\pm \frac{1}{2}$

Now,

$c+\lambda =\pm \frac{1}{2}$

Hence, this is the answer.