Quadratic equation $ 4ax^2 + 7x + b = 0$ $ 4ax^2 + 7x + b = 0$, $a,b \not= 0$.
How do I find the possible values of $a$ and $b$, and for each of the values of $a$ and $b$ to form the quadratic equations with roots $-2a$ and $\frac{1}{2}b$?
Sorry if this is basic, but I just started learning this and I don't seem to be able to solve it.
Edit: I'm meant to find the possible values of $a$ and $b$ first, I think. Then only for each of the possible values of $a$ and $b$, to form the quadratic equations with roots $-2a$ and $\frac{1}{2}b$. How do I start by finding the possible values of $a$ and $b$ first?
 A: Note that in any quadratic equation $ax^2+bx+c=0$ we have $$x_1+x_2=\frac{-b}{a}$$ and $$x_1.x_2=\frac{c}{a}$$ where in $x_1,x_2$ are the solutions of the equation. Clearly you want the equation to have two distict solutions, so $\Delta=b^2-4ac>0$ as well.
A: If you want $4ax^2+7x+b=0$ have the solutions exactly $x=-2p$ and $x=\frac12 q$, note that you want $4ax^2+7x+b$ to be a multiple of $(x+2p)(x-\frac12q)$. By comparing leading coefficients, you should have more precisely that $4ax^2+7x+b = 4a(x+2p)(x-\frac12q)$. Go ahaead, multiply out and compare the other coefficients.
A: The quadratic equation with roots $-2a$ and $\frac12b$ is $x^2-(\frac12b-2a)x+\frac12b(-2a)=0\implies x^2+(\frac{4a-b}2)x-ab=0$
This equation will be identical with the given one: $4ax^2+7x+b=0$ i.e., will have the same roots if the ratios of the coefficients of the different powers of $x$ are same i..e,  if and only if $$\frac{4a}1=\frac{7}{\frac{4a-b}2}=\frac b{-ab}$$
or $4a=\frac{14}{4a-b}=-\frac1a$ as $b\ne 0$
$\implies 4a=-\frac1a\implies 4a^2+1=0$
and $\frac{14}{4a-b}=-\frac1a\implies 14a=b-4a\implies b=18a$
