Let $f(x)=ax^2+bx+c$ and $g(x)=x^2+d$.
Now consider $f(x)+\lambda g(x)=0$
It is given that $f(x)+\lambda g(x)=0$ has identical roots and such $\lambda$ values are also identical.
1) if $d>0$ , show that $f(x)=ag(x)$
2) if $d<0$ , show that $f(x)$ and $g(x)$ has a common factor.
My attempt
1) $f(x)+\lambda g(x)=(a+\lambda)x^2+bx+(c+\lambda d)=0$
Then $$b^2-4(a+\lambda)(c+\lambda d)=0$$
So we get $$4d\lambda^2+4(a+cd)\lambda-(b^2-4ac)=0$$
Since $\lambda$ values are also identical, we get
$$(ad+c)^2+d(b^2-4ac)=0$$
$$(ad-c)^2+b^2d=0$$
Now if $d>0$ , then both $(ad-c)^2=0$ and $b^2d=0$.
Thus $b=0$ and $ad=c$
then we get that $f(x)=ag(x)$
How can I solve the second part
Thank you
For the second part, the solution in (1) holds good till the line $(*)$ below, $$\begin{aligned}(ad-c)^2+b^2d&=0 \qquad \qquad (*)\\ \implies (-c-aD)^2-b^2D&=0 \ (\text{where }D=-d>0)\\ \implies (c+aD)^2-(b\sqrt{D})^2&=0\\ \implies (c+aD-b\sqrt D)(c+aD+b\sqrt D)&=0 \qquad \qquad (1)\\ \implies f(-\sqrt D)f(\sqrt D)&=0\\ \implies \text{at least one of $\sqrt D$ and $-\sqrt D$}&\text{ is a root of $f(x)$}\end{aligned}$$ but $\sqrt D, - \sqrt D$ are precisely the solutions of the second quadratic $$g(x)=x^2+d=0 \implies x^2-D=0 \implies (x-\sqrt{D})(x+\sqrt{D})=0$$ Thus one of the numbers $\sqrt D, - \sqrt D$ is a root of both the quadratics $f(x),g(x)$, and by the Factor Theorem, one of the terms $$(x-\sqrt D) \ \ \text{ or } \ \ (x+\sqrt D) $$ divides both $f$ and $g$, hence, is a common factor of both.