Mathematics - Quadratic Equation Question with Solution | TestHub

Given $a{x}^2+bx+c=0$ $$\begin{array}{rl}& \Rightarrow a{x}^2=-bx-c\\ & \text{ Now, consider }\\ & y=4a{x}^2+3bx+2c\\ & =4[-bx-c]+3bx+2c\\ & =-4bx-4c+3bx+2c\\ & =-bx-2c\end{array}$$ Since, this curve intersects $x$-axis $\therefore$ put $y=0$, we get $$\begin{array}{rl}& -bx-2c=0\Rightarrow -bx=2c\\ & \Rightarrow x=\frac{-2c}{b}\end{array}$$ Thus, given curve intersects $x$-axis at exactly one point.