Mathematics - Quadratic Equation Question with Solution | TestHub

For a quadratic equation $a{x}^2+bx+c=0$ with $a\ne 0$, if its roots lie on opposite sides of a real number $k$, then $a\cdot f(k)<0$. In this equation, $a=1$ and the point is $p$. So, we need $f(p)<0$. Substituting $x=p$ into the equation, we get $p^2-2(p)+p^2-4p+3<0$, which simplifies to $2p^2-6p+3<0$. To find the values of $p$ for which this inequality holds, we find the roots of $2p^2-6p+3=0$. Using the quadratic formula, $p=\frac{-(-6)\pm \sqrt{(-6{)}^2-4(2)(3)}}{2(2)}=\frac{6\pm \sqrt{36-24}}{4}=\frac{6\pm \sqrt{12}}{4}=\frac{6\pm 2\sqrt{3}}{4}=\frac{3\pm \sqrt{3}}{2}$. Since the parabola $y=2p^2-6p+3$ opens upwards, $2p^2-6p+3<0$ when $p$ is between its roots. Therefore, $p\in \left(\frac{3-\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\right)$.