Two Quadratics with roots $\alpha,\beta$ and $\gamma, \delta$

Question

Given that $\alpha,\beta$ are the roots of $x^2+bx+c=0$ and $\gamma,\delta$ are the roots of $x^2+b_1x+c_1=0$ such that $\gamma <\alpha<\delta<\beta$

Then $(c-c_1)^2$ is Less than which of the following:

$A.$ $(b_1-b)(bc_1-b_1c)$

$B.$ $1$

$C.$ $(b-b_1)^2$

$D.$ $(c-c_1)(b_1c-b_1c_1)$

My try:

We have $(\alpha-\beta)^2=b^2-4c$ and $(\gamma-\delta)^2=b_1^2-4c_1$

Then $$4(c-c_1)=(b^2-b_1^2)-(\alpha-\beta)^2+(\gamma-\delta)^2$$

Any clue here?

Answer 1

Hint: Use numerical values $1<2<3<4$ to rule out B, C, and D.

To prove A, just convert all expressions as polynomials in $\alpha,\beta,\gamma,\delta$. Taking difference and factoring yields the final result. More precisely it boils down to $$(\alpha-\gamma)(\delta-\alpha)(\beta-\gamma)(\beta-\delta)>0.$$