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

Question

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

I can't figure out a way to solve this question. A hint would be great.

Answer 1

Assuming $c$ is real.

Since $\lambda^2-4 <0$, therefore both roots are complex. Also since both polynomials are with real coefficients, so the complex roots occur in conjugate pairs. This means that both polynomials will share both complex roots. Thus the quadratic polynomial will divide the cubic polynomial. Suppose $\alpha, \beta$ are the roots of the quadratic polynomial, then using Viete's relations, the third root $r$ of the cubic will be $$r=-(\alpha+\beta)=\lambda$$ Again using Viete's relations we get, $$r\alpha\beta=\lambda(1)=-c$$ Therefore $$\lambda+c=0.$$

Answer 2

If $\lambda$ is in $(-2,2)$ (not to be confused with $[-2,2]$), then the roots of $x^2+\lambda x+1=0$ are complex because $x^2+\lambda x +1=0$ only has real solutions when $\lambda=2$ and $\lambda = -2$.

Let $a$ be a shared solution of $x^2+\lambda x+1=0$ and $4x^3+3x+2c=0$. Since $a$ is complex, $4a^3+3a+2c=0$ gives us $c=-\frac{4a^3+3a}{2}$. Thus, $c$ is complex, so $\lambda + c$ is complex.