Question

If $Î\pm ,β$ and $γ$ are the roots of the equation $x^3−13x^2+15x+189=0$ and one root exceeds the other by $2,$ then the value of $âˆ\pounds Î\pm âˆ\pounds +âˆ\pounds βâˆ\pounds +âˆ\pounds γâˆ\pounds$ is equal to

• A. $23$
• B. $17$
• C. $13$
• D. $19$

Answer 1

Answer: (D)

Let, the roots be $Î\pm ,β,Î\pm +2.$
$S_1=Î\pm +β+Î\pm +2=2Î\pm +β+2=13⇒2Î\pm +β=11⇒β=11−2Î\pm$
$S_2=Î\pm β+β\left(Î\pm +2\right)+\left(Î\pm +2\right)Î\pm =15$
$⇒β\left(Î\pm +Î\pm +2\right)+Î\pm \left(Î\pm +2\right)=15$
$⇒\left(11−2Î\pm \right)\left(2Î\pm +2\right)+Î\pm \left(Î\pm +2\right)=15$
$⇒22Î\pm +22−4{Î\pm }^2−4Î\pm +{Î\pm }^2+2Î\pm =15$
$⇒3{\left(Î\pm \right)}^2−20Î\pm −7=0⇒\left(Î\pm −7\right)\left(3Î\pm +1\right)=0$
$⇒Î\pm =7$ or $−\frac{1}{3}.$
$Î\pm =7,β=11−2Î\pm =11−14=−3,γ=Î\pm +2=9$
$Î\pm =−\frac{1}{3},β=11−2Î\pm =11+\frac{2}{3}=\frac{35}{3},γ=Î\pm +2=\frac{5}{3}.$
Since, $Î\pm βγ=−189$ , hence we will take the first case.
$âˆ\pounds Î\pm âˆ\pounds +âˆ\pounds βâˆ\pounds +âˆ\pounds γâˆ\pounds =âˆ\pounds 7âˆ\pounds +âˆ\pounds −3âˆ\pounds +âˆ\pounds 9âˆ\pounds =19$