If $a,b,c$ all real numbers and $1$ is root of $ax^2 + bx + c = 0$, then $y = 4ax^2 + 3bx + 2c$ , a is not 0 intersects x-axis at how many points?
I could infer that $a + b + c = 0$ and $4ax^2 + 3bx + 2c = 0$. But what else should I do to solve the problem?
We have the discriminant of $y=4ax^2+3bx+2c $ as: $$\Delta = B^2-4AC $$ $$=(3b)^2-4 (4a)(2c) $$ $$=9b^2-32ac $$ $$=9 (-a-c)^2-32ac $$ $$=9 (a-c)^2+ 4ac $$ Hence, both roots are real and will cut the $x$-axis at $\boxed {2\text { points }} $. Hope it helps.