Quadratic Equation Roots and Coefficients: Solving for Unknowns

[rtwikia]

$\alpha$ and $\beta$ are the roots of the equation $2{x}^{2}-5x+c=0$. If $4\alpha-2\beta=7$, find the value of $c$.

I did the following:

$\alpha+\beta=-\frac{-5}{2}=\frac{5}{2}$
$\alpha\beta=\frac{c}{2}$

$\frac{c}{2}=\frac{7+2\beta}{4}\cdot\frac{-7+4\alpha}{2}$
$$=\frac{7+2\beta}{4}\cdot\frac{-14+8\alpha}{4}$$
$8c=(7+2\beta)(-14+8\alpha)$
$$=(2\beta+7)(8\alpha-14)$$
$$=16\alpha\beta-28\beta+56\alpha-98$$
$4c=8\alpha\beta-14\beta+28\alpha-49$ which re-factorizes into $(4\alpha-7)(2\beta+7)$

I could not go any further. Can anyone help me

 

[I like Serena]

$\alpha$ and $\beta$ are the roots of the equation $2{x}^{2}-5x+c=0$. If $4\alpha-2\beta=7$, find the value of $c$.

I did the following:

$\alpha+\beta=-\frac{-5}{2}=\frac{5}{2}$
$\alpha\beta=\frac{c}{2}$

$\frac{c}{2}=\frac{7+2\beta}{4}\cdot\frac{-7+4\alpha}{2}$
$$=\frac{7+2\beta}{4}\cdot\frac{-14+8\alpha}{4}$$
$8c=(7+2\beta)(-14+8\alpha)$
$$=(2\beta+7)(8\alpha-14)$$
$$=16\alpha\beta-28\beta+56\alpha-98$$
$4c=8\alpha\beta-14\beta+28\alpha-49$ which re-factorizes into $(4\alpha-7)(2\beta+7)$

I could not go any further. Can anyone help me

Hi rtwikia! Welcome to MHB! (Smile)

We have 3 equations with 3 unknowns:
$$\alpha+\beta=\frac{5}{2}$$
$$\alpha\beta=\frac{c}{2}$$
$$4\alpha-2\beta=7$$

That means we can get a solution by eliminating one variable after another... (Thinking)

 

[rtwikia]

...
That means we can get a solution by eliminating one variable after another... (Thinking)

Yes! I got it!(Rock)

$\beta=\frac{1}{2}\implies\alpha=2\implies c=2$

Thanks for your help!(Rock)