Question

If $\alpha$ and $\beta$ are the roots of $5x^2-px+1=0$ and $\alpha -\beta =1$, then find p.

• A. $p=\pm 3\sqrt{5}$
• B. $p=\pm 2\sqrt{5}$
• C. $p=\pm 3\sqrt{2}$
• D. None of these

Answer 1

The correct option is A $p=\pm 3\sqrt{5}$

$\Rightarrow$ $\alpha$ and $\beta$ re the roots of $5x^2-px+1=0$

$\Rightarrow$ Here, $a=5,b=-p,c=1$

$\Rightarrow$ $\alpha -\beta =1$ ----- ( 1 ) [ Given ]

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{1}{5}$ ----- ( 2 )

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-(-p)}{5}=\frac{p}{5}$ -----( 3 )

$\Rightarrow$ $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow$ ${\left(\frac{p}{5}\right)}^2={\alpha }^2+{\beta }^2+2\times \frac{1}{5}$ [ Using ( 2 ) and ( 3 ) ]

$\Rightarrow$ ${\alpha }^2+{\beta }^2=\frac{p^2}{25}-\frac{2}{5}$ ----- ( 4 )

$\Rightarrow$ $(\alpha -\beta {)}^2={\alpha }^2+{\beta }^2-2\alpha \beta$

$\Rightarrow$ $(1{)}^2=\frac{p^2}{25}-\frac{2}{5}-\frac{2}{5}$ [ Using ( 1 ),( 2 ) and ( 4 ) ]

$\Rightarrow$ $1=\frac{p^2}{25}-\frac{4}{5}$

$\Rightarrow$ $\frac{p^2}{25}=1-\frac{4}{5}$

$\Rightarrow$ $p^2=\frac{9}{5}\times 25$

$\Rightarrow$ $p^2=45$

$\therefore$ $p=\pm 3\sqrt{5}$