Question

If $x=-3$ and $x=\frac{2}{3}$ are solutions of quadratic equation $m{x}^2+7x+n=0,$ find the values of $m$ and $n$.

Answer 1

Step 1 : Form linear equations in $m$ and $n$
For $x=-3$ and $x=\frac{2}{3}$ to be solution of the given quadratic equation it should satisfy the equation $m{x}^2+7x+n=0$
Substitute $x=-3$ in the given equation
$m(-3{)}^2+7(-3)+n=0$
$9m-21+n=0$
$\Rightarrow 9m+n=21$ ....(i)
Substitute $x=\frac{2}{3}$ in the given equation
$m{\left(\frac{2}{3}\right)}^2+7\left(\frac{2}{3}\right)+n=0$
$\frac{4m}{9}+\frac{14}{3}+n=0$
$4m+42+9n=0$
$\Rightarrow 4m+9n=-42$ ......(ii)

Step 2: Solve linear equations in $m$ and $n$
We have two equations
$9m+n=21$ ..... (i)
$4m+9n=-42$ ....(ii)
Multiply (i) by $9$
$(9m+n=21)\times 9$
$81m+9n=189$ .... (iii)
Solve equation (iii) and (ii)
$77m=231$
$m=\frac{231}{77}=\frac{21}{7}=3$
Put $m=3$ in equation (i), we get $n=-6$.
Hence, the values of $m$ and $n$ are $3$ and $-6$, respectively.