Question

If the product of two roots of $x^3+k{x}^2+31x-30=0$ is $15$, then $k=$

• A. $2$
• B. $-2$
• C. $-10$
• D. $10$

Answer 1

The correct option is C $-10$

Let the roots of given equation $x^3+k{x}^2+31x-30=0$ is $p,q,r$

by given condition,

$p\times q=15$

Relation betwen roots and coefficients are

Product of the roots$=\frac{-d}{a}$

$\therefore p\times q\times r=\frac{-d}{a}=\frac{-(-30)}{1}=30$

$15r=30$

$r=2$,

Sum of products of the roots taken two at a time$=\frac{c}{a}$

$\therefore p\times q+p\times r+q\times r=\frac{c}{a}=\frac{31}{1}=31$

$15+r(p+q)=31$

$r(p+q)=16$

$p+q=16/2=8$ and also

$pq=15$

So $p+q+r=\frac{-b}{a}=\frac{-k}{1}=-k$

$(p+q)+r=-k$

$8+2=-k$

$k=-10$