Question

$\frac{2}{3}$ and 1 are the solutions of equations $m{x}^2+nx+1=0$. Find the value of m and n

Answer 1

Step 1: Solving for $\frac{2}{3}$.
$\frac{2}{3}$ is solution of $m{x}^2+nx+1=0$

Therefore,

$m\times x^2+n\times x+1=0$

$\Rightarrow m\times {\left(\frac{2}{3}\right)}^2+n\times \frac{2}{3}+1=0$

$\Rightarrow m\times \frac{4}{9}+n\times \frac{2}{3}+1=0$

$\Rightarrow \frac{4m}{9}+\frac{2n}{3}+1=0$

$\Rightarrow \frac{4m+2n\times 3+1\times 9}{9}=0$

$\Rightarrow \frac{4m+6n+9}{9}=0$

$\Rightarrow 4m+6n+9=0...\left(i\right)$

Step 2: Solving for 1.

1 is solution of $m{x}^2+nx+1=0$

Therefore,

$m\times x^2+n\times x+1=0$

$\Rightarrow m\times (1{)}^2+n\times 1+1=0$

$\Rightarrow m+n+1=0$ $...\left(ii\right)$

Step 3: Solving the value of m and n.

$4m+6n+9=0...\left(i\right)$

$m+n+1=0...\left(ii\right)$

From equation (ii), we get
$m=-(n+1)$
Putting the value of m in equation (i), we get
$4[-(n+1)+6n+9=0$
$\Rightarrow -4n-4+6n+9=0$
$\Rightarrow 6n-4n+9-4=0$
$\Rightarrow 2n=-5$
$\Rightarrow n=-\frac{5}{2}=-2\frac{1}{2}$

Putting the value of n in equation (ii), we get
$m+(-\frac{5}{2})+1=0$
$\Rightarrow m=\frac{5}{2}-1$
$\Rightarrow m=\frac{5-2}{2}=\frac{3}{2}=1\frac{1}{2}$
Hence.
$m=1\frac{1}{2}$& $n=-2\frac{1}{2}$.