also $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta$
$\Rightarrow \quad \alpha \beta=-\frac{7}{4}$
required quadratic equation is
$ x^{2}-x(3)-\frac{7}{4}=0 $
$ \Rightarrow 4 x^{2}-12 x-7=0$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ The equation of circle $\mathrm{C}$ is
$(A)$ $(x-2 \sqrt{3})^2+(y-1)^2=1$
$(B)$ $(x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1$
$(C)$ $(x-\sqrt{3})^2+(y+1)^2=1$
$(D)$ $(x-\sqrt{3})^2+(y-1)^2=1$
$2.$ Points $E$ and $F$ are given by
$(A)$ $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$
$(B)$ $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right),(\sqrt{3}, 0)$
$(C)$ $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$(D)$ $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$3.$ Equation of the sides $Q R, R P$ are
$(A)$ $y=\frac{2}{\sqrt{3}} x+1, y=-\frac{2}{\sqrt{3}} x-1$
$(B)$ $y=\frac{1}{\sqrt{3}} x, y=0$
$(C)$ $y=\frac{\sqrt{3}}{2} x+1, y=-\frac{\sqrt{3}}{2} x-1$
$(D)$ $y=\sqrt{3} x, y=0$
Give the answer question $1,2$ and $3.$