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?
 A: So you have $a+b+c=0$ and the number of intersections between $y$ and the $x$ axis depends on the sign of the discriminant in this case $\Delta=(3b)^2-4(4a)(2c)$. Inserting $b=-a-c$ we get 
$$\Delta=(3(-a-c))^2-4(4a)(2c)\\
=9(a+c)^2-4(4a)(2c)\\
=9a^2+9c^2-14 a c$$
which attains a minimum value of 0 at $a=0,c=0$ and since you've excluded $a=0$ $\Delta$ is strictly positive so there are always two intersections with the $x$ axis.
A: Since $c=-a-b$, the discriminant $\Delta$ is
$$\begin{align}\Delta&=(3b)^2-4\cdot 4a\cdot 2c\\&=9b^2-32a(-a-b)\\&=32a^2+32ab+9b^2\\&=32a^2+32ab+8b^2+b^2\\&=8(4a^2+4ab+b^2)+b^2\\&=8(2a+b)^2+b^2\\&\gt 0\end{align}$$
because $\Delta=0\iff 2a+b=b=0\iff a=b=0$, which contradicts $a\not=0$.
A: Case 1: $\; c=0 \implies b=-a$
$4ax^2+3bx+2c \equiv ax(4x-3)$ which has zeroes of $0$ and $\frac{3}{4}$.

Case 2: $\;ax^2+bx+c\equiv a(x-1)^{2} \implies a=c=-\frac{b}{2}$
$4ax^2+3bx+2c \equiv 2a(2x-1)(x-1)$ which has zeroes of $\frac{1}{2}$ and $1$.

Case 3: $\; a\xi^2+b\xi+c=0 \:$ where $\: \xi \neq 0,1$
Considering $f(x)=x^2(ax^2+bx+c)$ with $a+b+c=0$.
$f(0)=f(1)=f(\xi)=0$
By Rolle's Theorem,
$\exists \lambda, \mu \in (\min[0,\xi],\max[1,\xi])$ such that $f'(\lambda)=f'(\mu)=0$ where $\lambda$, $\mu$, $0$, $1$, $\xi$ are all distinct.
$ \implies
\left \{
  \begin{eqnarray*}
    4a\lambda^{3}+3b\lambda^{2}+2c\lambda &=& 0 \\
    4a\mu^{3}+3b\mu^{2}+2c\mu &=& 0
  \end{eqnarray*}
\right.$
That is $\lambda,\mu$ are the zeroes of $4ax^{2}+3bx+2c$

Hence $\boldsymbol{4ax^{2}+3bx+2c}$ has two distinct zeroes.
A: $a = -(b+c)$
The discriminant of the second quadratic is therefore:
$9b^2 - 32ac = 9b^2 + 32bc + 32c^2 = f(b,c)$
where we defined a new function $f(b,c)$
You can easily show that $f(b,c)$ is always positive by considering its own discriminant (which is always negative) and noting that at least one of $b$ or $c$ is nonzero.
Hence the second quadratic has exactly two real roots.
A: 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.
A: If $1$ is a root of $ax^2 + bx + c = 0$ Then 
\begin{align}
    ax^2 + bx + c
        &= a(x-1)(x - \lambda) \\
        &= ax^2 - a(1 + \lambda)x + a\lambda \\
\end{align}
So $b = -a(1+\lambda)$ and $c = a\lambda$.
The descriminant of $4ax^2 + 3bx + 2c$ is
\begin{align}
    9b^2 - 32ac
        &= 9a^2(1+\lambda)^2 -32a^2\lambda \\
        &= a^2(9\lambda^2-14\lambda+9) \\
        &= 9a^2\left[\left(x- \dfrac 79 \right)^2 + \dfrac{32}{9}\right]
\end{align}
Which is positive.
Hence the polynomial has two real roots.
