Question

If $x=2$ and $x=3$ roots of the equation $3x^2-2mx+2n=0$, then the values of m and n are

• A. $m=\frac{15}{2},n=9$
• B. $m=-\frac{15}{2},n=-9$
• C. $m=\frac{15}{2},n=-9$
• D. $m=-\frac{15}{2},n=9$

Answer 1

The correct option is A $m=\frac{15}{2},n=9$

Let $f\left(x\right)=3x^2-2mx+2n=0$

$x=2$ and $x=3$ are the roots of $f\left(x\right)$

Thus,

$f\left(2\right)=3{\left(2\right)}^2-2m\left(2\right)+2n=0$

$=>\Rightarrow 12-4m+2n=0$

$\Rightarrow 2m-n-6=0$-----------------(i)

$f\left(3\right)=3{\left(3\right)}^2-2m\left(3\right)+2n=0$

$\Rightarrow 27-6m+2n=0$

$\Rightarrow 6m-2n-27=0$--------------(ii)

Solving, (i) and (ii), we get

$\frac{m}{27-12}=\frac{n}{-36+54}=\frac{1}{-4+6}$

$=>\frac{m}{15}=\frac{n}{18}=\frac{1}{2}$

$=>m=\frac{15}{2},n=\frac{18}{2}=9$