Discussion Forum

Let   $\alpha$   and $\beta$ be the roots of equation $px^{2}+qx+r=0$ , $p\neq0$ . If p,q and r in AP and  $\frac{1}{\alpha}+\frac{1}{\beta}=4$  , then the value of   $|\alpha-\beta|$ is 

Answer:

Explanation:

If    $ax^{2}+bx+c=0$ ,  has roots  $\alpha$ and $\beta$   , then 

  $\alpha+\beta=\frac{-b}{a}$    and    $\alpha\beta=\frac{c}{a}$ . Find the valuesw of   $\alpha+\beta$  and $\alpha \beta$ and then put in     $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta$    to get required value.

  Given   $\alpha$   and $\beta$    are roots of   

                          $px^{2}+qx+r=0$ , $p\neq0$ 

  $\therefore$    $\alpha+\beta$  = $\frac{-q}{p}$  ,   $\alpha \beta$=   $\frac{r}{p}$    ......(i)

Since p,q and r are in A.P

$\therefore$      2q=p+r        ................(ii)

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

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

$\Rightarrow$       $\alpha+\beta=4\alpha\beta$

$\Rightarrow$   $\frac{-q}{p}=\frac{4r}{p}$  [ from Eq.(i)]

$\Rightarrow$     q=-4r

 On putting the value of q in Eq(ii) , we get  $\Rightarrow$  2(-4r)=p+r  $\Rightarrow$  p=-9r

Now  $\alpha+\beta=\frac{-q}{p}=\frac{4r}{p}=\frac{4r}{-9r}=-\frac{4}{9}$    and $\alpha\beta=\frac{r}{p}=\frac{r}{-9r}=\frac{1}{-9}$

$\therefore$    $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta=\frac{16}{81}+\frac{4}{9}=\frac{16+36}{81}$

$\Rightarrow$    $(\alpha-\beta)^{2}=\frac{52}{81}\Rightarrow|\alpha-\beta|=\frac{2}{9}\sqrt{31}$