Browse Questions

# Find the value of P such that the difference of the roots of the equation $\;x^{2}-Px+8=0\;$ is $2$

$(a)\;\pm3\qquad(b)\;\pm6\qquad(c)\;\pm2\qquad(d)\;4$

Answer : $\;\pm6$
Explanation :
Let $\;\alpha \;, \beta\;$ be the roots of the equation $\;x^{2}-Px+8=0\;$
Therefore , $\;\alpha+\beta=P\;$ and $\; \alpha\;.\beta = 8$
Now , $\;\alpha - \beta = \pm \sqrt{(\alpha + \beta)^{2} - 4 \alpha \beta}$
Therefore , $\;2= \pm \sqrt{ P^{2} -32}$
$P^{2}-32=4$
i.e , $\;P=\pm 6\;.$