<s>[INST] 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?
 [/INST] 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.
</s>