$\alpha, \beta$ are the roots of the equation $(a-2)x^2-(5-a)x-5=0$. Find $a$ if $|\alpha - \beta|=2 \sqrt {6}$

Question

$\alpha, \beta$ are the roots of the equation $(a-2)x^2-(5-a)x-5=0$. Find $a$ if $|\alpha - \beta|=2 \sqrt {6}$

I cannot understand how to proceed with this problem. Please show me the logic behind this problem.

Answer 1

By the Vieta formulas,

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

$$\alpha\beta=-\frac5{a-2}.$$

Then

$$|\alpha-\beta|=\sqrt{(a+b)^2-4\alpha\beta}=\frac{\sqrt{a^2+10a-15}}{|a-2|}=2\sqrt 6$$

which you can solve for $a$ (two solutions).

Answer 2

Note that by Vieta's formulas, we have $$\alpha+\beta=-\frac{-(5-a)}{a-2}=-\frac{a-5}{a-2}=-1+\frac3{a-2}$$ and $$\alpha\beta=-\frac5{a-2}$$ so $$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta.$$ Hence all you need to do is solve $$\sqrt{(\alpha+\beta)^2-4\alpha\beta}=2\sqrt6$$ for $a$.

Answer 3

Hint:

The condition is equivalent to $(\alpha-\beta)^2=24$, and $$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta,$$ a symmetric function of $\alpha$ and $\beta$.