I have an issue with the following problem:
Two quadratic equations have real roots $\alpha$ and $\beta$ such that $$\alpha - \beta = 3$$ and $$\alpha \beta = 2(\alpha + \beta).$$ Find the two possible quadratic equations that satisfy these conditions.
Since $\alpha$ is larger than $\beta$, in the general solution to the quadratic equation, familiarly:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$\alpha$ needs to be the larger solution so it must be of the form:
$$\alpha = \frac{-b - \sqrt{b^2-4ac}}{2a}$$
since it's the bigger root, and $\beta$ is therefore:
$$\beta = \frac{-b + \sqrt{b^2-4ac}}{2a}$$
However, I need to find the two quadratic equations that have these roots.
If I play around a bit with the properties of the roots, I can work out some things like:
$$\alpha - \beta = 3$$
$$\frac{1}{2a}\ ((-b - \sqrt{b^2-4ac}) - (-b + \sqrt{b^2-4ac}) = 3$$
$$ - 2\sqrt{b^2-4ac} = 6a$$
$$4b^2 - 16ac = 36a^2$$
$$b = \frac{\sqrt{16a(2a - c)}}{2}$$
And the next property:
$$\alpha \beta = 2(\alpha + \beta)$$
$$\frac{-b - \sqrt{b^2-4ac}}{2a} \bullet \frac{-b + \sqrt{b^2-4ac}}{2a} = 2(\frac{1}{2a}(-b - \sqrt{b^2-4ac}+ -b + \sqrt{b^2-4ac}))$$
$$\frac{4c}{4a}=\frac{-2b}{a}$$
$$4c=\sqrt{16a(2a - c)}$$
$$16c^2 = 32a^2 - 16ac$$
$$0= 2a^2-ac -c^2$$
$$(a-c)(2a+c)=0$$
$$a= c$$ $$2a = -c$$
I'm going to assume these will represent both quadratics, so I'll use $a = c$ first.
$$b = \frac{\sqrt{16a(2a - c)}}{2}$$
$$b = \frac{\sqrt{16c(2c - c)}}{2}$$
$$b = \frac{\sqrt{16c^2}}{2}$$ $$b= \pm\ 2c$$
I'll use $b = 2c$ for equation one.
$$\frac{1}{2a}\ ((-b - \sqrt{b^2-4ac}) - (-b + \sqrt{b^2-4ac}) = 3$$
$$\frac{1}{2c}\ ((-2c - \sqrt{{4c}^2-4c^2}) - (-2c + \sqrt{{4c}^2-4c^2}) = 3$$
$$c=0$$
Obviously, I blew it, since that implies $a = b = c$. I'm not sure if what I did wrong was some technical errors or the wrong approach in itself. I'd appreciate any guidance on this.