If $\alpha, \beta$ are the roots of $x^2+b x+c=0, \gamma, \delta$ are the roots of $x^2+b_1 x+c_1=0$ and $\gamma

Question

If $\alpha, \beta$ are the roots of $x^2+b x+c=0, \gamma, \delta$ are the roots of $x^2+b_1 x+c_1=0$ and $\gamma < \alpha < \delta < \beta$, then $\left(c-c_1\right)^2 < $

Aicharya · Accepted Answer

According to given informations after drawing figure.

For $x$-coordinate of point $P$, on subtracting given quadratic equations, we get $\begin{aligned} x^2+b_1 x+c_1 & =0 \\ x^2+b x+c & =0 \end{aligned}$ $\begin{aligned} & \left(b_1-b\right) x+\left(c_1-c\right)=0 \\ & \Rightarrow x=\left(\frac{c-c_1}{b_1-b}\right) \\ & \end{aligned}$ Now, with respect to quadratic expression $\begin{gathered} f(x)=x^2+b x+c \\ f\left(x=\frac{c-c_1}{b_1-b}\right) < 0 \\ \Rightarrow\left(\frac{c-c_1}{b_1-b}\right)^2+b\left(\frac{c-c_1}{b_1-b}\right)+c < 0 \\ \Rightarrow\left(c-c_1\right)^2 < b\left(c_1-c\right)\left(b_1-b\right)-c\left(b_1-b\right)^2 \\ \Rightarrow\left(c-c_1\right)^2 < \left(b_1-b\right)\left[b c_1-b c-c b_1+c b\right] \\ \Rightarrow\left(c-c_1\right)^2 < \left(b_1-b\right)\left(b c_1-c b_1\right) \end{gathered}$ Hence, option (1) is correct.

If \(\alpha, \beta\) are the roots of \(x^2+b x+c=0, \gamma, \delta\) are the roots of \(x^2+b_1 x+c_1=0\)…

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