If alpha, beta are the roots of x^2+bx+c=0,ga | vedclass

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(A) Let $f(x) = x^2+bx+c$ and $g(x) = x^2+b_1x+c_1$. The intersection point $P$ of the two parabolas is found by solving $f(x) = g(x)$.
$x^2+bx+c = x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $f(x) = x^2+bx+c$ and $g(x) = x^2+b_1x+c_1$. The intersection point $P$ of the two parabolas is found by solving $f(x) = g(x)$.
$x^2+bx+c = x^2+b_1x+c_1$
$(b-b_1)x = c_1-c$
$x = \frac{c_1-c}{b-b_1} = \frac{c-c_1}{b_1-b}$.
Since $\gamma < \alpha < \delta < \beta$,the point of intersection $P$ lies below the $x$-axis,meaning $f(x) < 0$ at this $x$-coordinate.
$f\left(\frac{c-c_1}{b_1-b}\right) < 0$
$\left(\frac{c-c_1}{b_1-b}\right)^2 + b\left(\frac{c-c_1}{b_1-b}\right) + c < 0$
Multiplying by $(b_1-b)^2$ (which is positive):
$(c-c_1)^2 + b(c-c_1)(b_1-b) + c(b_1-b)^2 < 0$
$(c-c_1)^2 < -b(c-c_1)(b_1-b) - c(b_1-b)^2$
$(c-c_1)^2 < (b_1-b)[-b(c-c_1) - c(b_1-b)]$
$(c-c_1)^2 < (b_1-b)[-bc+bc_1-cb_1+cb]$
$(c-c_1)^2 < (b_1-b)(bc_1-b_1c)$.
Thus,option $A$ is correct.

If $\alpha, \beta$ are the roots of $x^2+bx+c=0$,$\gamma, \delta$ are the roots of $x^2+b_1x+c_1=0$ and $\gamma < \alpha < \delta < \beta$,then $(c-c_1)^2 < $

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