Find the roots of quadratic polynomial given one root of another quadratic polynomial? if $a,b,c$ are Real numbers and $1$ is a root of equation $ax^2+bx+c=0$ then curve $y = 4ax^2+3bx+2c$ , (a is not zero) intersects $x$ axis at how many points?
I get a relation $a+b+c = 0$ I tried substituting this in given polynomial and tried finding its discriminant but cannot figure out how to find whether it is positive, negative or zero? can someone help?

You found that $a+b+c=0$, which means that $c=-a-b$, so we have $y=4ax^2+3bx-(2a+2b)$. The discriminant is as follows:
$$(3b)^2-4(4a)(-(2a+2b))=9b^2+16a(2a+2b)=9b^2+32a^2+32ab$$
Now, if we plug in $a=1$ and $b=1$ into the discriminant, we clearly get a positive number, meaning there are two solutions. We need to prove that we get a positive number for all $b \in \Bbb{R}$ and all $a \in \Bbb{R}-\{0\}$.
In order to do this, let's try to express the discriminant as the sum of two squares. We can do this by completing the square with $32a^2+32ab$. First, factor out a $32$ to get $32(a^2+ab)$. Then, divide the coefficient of $ab$ by $2$ to get $\frac 1 2$. This will be the coefficient of $b$ in the monomial squared. Then, square that to get $\frac 1 4$ to get the coefficient of $b^2$. Finally, add $\frac 1 4 b^2$ inside the parentheses to get the following:
$$32(a^2+ab+\frac 1 4 b^2)=32a^2+32ab+8b^2=32\left(a+\frac 1 2 b\right)^2$$
We now know that $32a^2+32ab+8b^2=32\left(a+\frac 1 2 b\right)^2$. However, the discriminant is $32a^2+32ab+9b^2$ which can be expressed as the sum of two squares as follows:
$$32a^2+32ab+9b^2=32a^2+32ab+8b^2+b^2=32\left(a+\frac 1 2 b\right)^2+b^2$$
Thus, our discriminant is the sum of two squares with one of the squares being multiplied by a positive number, which means that the discriminant 
is always non-negative. The only case where it is $0$ is if $b=0$ and $a+\frac 1 2 b=0$. Combining these two equations finds that $a=0$, which you said is not under the constraints. Therefore, the discriminant is always positive.
