Let $α$ and $β$ be the roots of equation $px^2+qx+r=0,p≠0$

Question

Let $α$ and $β$ be the roots of equation $px^2+qx+r=0,p≠0$, If $p,q,r$ are in A.P and $\dfrac{1}{α}+\dfrac{1}{β}=4$, then the value of $|α−β|$ is $:$

1. $\dfrac{\sqrt{61}}{9} $
2. $\dfrac{2\sqrt{17}}{9}$
3. $\dfrac{\sqrt{34}}{9}$
4. $\dfrac{2\sqrt{13}}{9}$

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$\text{Somewhere it explained as:}$

$$\frac{1}{\alpha}+\frac{1}{\beta}=4$$

$$2q = p + r$$

$$\Rightarrow \space\space-2(\alpha+\beta)=1+\alpha\beta$$

$$\Rightarrow-2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\frac{1}{\alpha\beta}+1\cdots(1)$$

$$\Rightarrow\frac{1}{\alpha\beta}=-9$$

$\text{Equation having roots $α$ and $β$ is:}$ $$9x^2\space +\space 4x\space -1=0$$

$$\alpha,\space \beta=\frac{-4\underline{+}\sqrt{16+36}}{2\times 9}$$

$$\left|\alpha-\beta\right|=\frac{2\sqrt{13}}{9}$$

$\text{As I know:} $$$x^2-(\alpha+\beta)+(\alpha\times\beta)=0\cdots(2)$$

$\text{My question is:}$

Can you explain in formal/alternative way, please?

Answer 1

Let $\alpha$ and $\beta$ be the roots of the equation $ax^2+bx+c=0$. By the usual formula, the difference of the roots is $\frac{\sqrt{b^2-4ac}}{a}$. Hence it's enough for us to find the correct $a,b,c$ for which the roots are $\alpha$ and $\beta$.

But here's the trick: So we know that $\frac{1}{\alpha} + \frac{1}{\beta} = 4$. That means that if we look at the equation $2q = p + r$, we get that $\frac{1}{\alpha \beta} = -9$. So if $ex^2+fx+g=0$is the equation satisfied by $\frac{1}{\alpha}$ and $\frac{1}{\beta}$, then $e=1$, $f$ is the negative of the sum which is $-4$, and $g$ is the product, which is $-9$. Putting together, this gives us $x^2-4x-9=0$ is the equation satisfied by $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. To get the equation satisfied by $\alpha$ and $\beta$, all we do is invert the $x$ by making it $\frac{1}{x}$, and then that gives us $9x^2+4x-1=0$. Finally, using the formula we know above of the difference of the roots, we get the answer as $\frac{2 \sqrt{13}}{9}$.