Question

Solve the given quadratic equation

$m^2=4+5m$, then $m=\frac{5\pm \sqrt{41}}{2}$
If true then enter $1$ and if false then enter $0$

Answer 1

Given, $m^2=4+5m\Rightarrow m^2-5m-4=0$

Compare the given equation with $a{x}^2+bx+c=0$

Since, $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$

In given quadratic equation $a=1,b=-5,c=-4$

$\therefore m=\frac{-(-5)+\sqrt{(-5{)}^2-4\times 1\times (-4)}}{2a}$

and

$m=\frac{-(-5)-\sqrt{(-5{)}^2-4\times 1\times (-4)}}{2a}$

$\therefore m=\frac{5+\sqrt{41}}{2}$ and $m=\frac{5-\sqrt{41}}{2}$
$\therefore m=\frac{5\pm \sqrt{41}}{2}$
Thus given statement is true