Question

If $n^2+3n-4=0$, then $n=1,-4$
If true then enter $1$ and if false then enter $0$.

Answer 1

Given equation is $n^2+3n-4=0$

$\therefore n^2+3n=4$

Third term $=$ ${(\frac{1}{2}\times \text{coefficient of x})}^2$ $=$ ${(\frac{1}{2}\times 3)}^2=\frac{9}{4}$

Adding $\frac{9}{4}$ on both sides, we get

$n^2+3n+\frac{9}{4}=4+\frac{9}{4}$

$\Rightarrow {(n+\frac{3}{2})}^2=\frac{25}{4}$

$\Rightarrow (n+\frac{3}{2})$ = + $\frac{5}{2}$

$\Rightarrow n=\frac{5}{2}-\frac{3}{2}$ OR $n=-\frac{5}{2}-\frac{3}{2}$

$\Rightarrow n=1$ OR $n=-4$