Question

If $2$ and $3$ are the roots of the equation $2x^3+m{x}^2-13x+n=0,$ the third root of this equation is

• A. $-5$
• B. $30$
• C. $-\frac{5}{2}$
• D. $-15$

Answer 1

Answer: (C)

$-\frac{5}{2}$

$2x^3+m{x}^2-13x+n=0$

Substituting $x=2$

$2\times 2^3+m\times 2^2-13\left(2\right)+n=0$

$16+4m-26+n=0$

$4m+n=10$ (1)

Substituting $x=3$

$2\times 3^3+m\times 3^2-13\left(3\right)+n=0$

$54+9m-39+n=0$

$9m+n=-15$ (2)

Subtracting (1) from (2)

$5m=-25$

$m=-5$ (3)

Substituting (3) in (1)

$4(-5)+n=10$

$-20+n=10$

$n=30$

Let the third root be $a$

Now product of roots $=-\frac{n}{2}$

$2\times 3\times a=-\frac{30}{2}$

$6\times a=-15$

$a=\frac{-5}{2}$

Hence the correct option is (C).