Question

If the roots of $x^3-12x^2+39x+k=0$ are in A.P, then $k=$

• A. $28$
• B. $-28$
• C. $18$
• D. $-18$

Answer 1

The correct option is B $-28$

As roots are in A.P. then
let $a-d,a,a+d$ are the roots of $x^3-12x^2+39x+k=0$
$S_1=a-d+a+a+d=12\Rightarrow a=4$
Substituting this in equation we get
${\left(4\right)}^3-12{\left(4\right)}^2+39\left(4\right)+k=0\Rightarrow 64-96+156+k=0\Rightarrow k=-28$