Question

Find the value of m so that the quadratic equation $mx(x-7)+49=0$ has two roots.

Answer 1

Given that the quadratic equation

$mx(x-7)+49=0$

find the value of m

Then solve

$\Rightarrow mx(x-7)+49=0$

$\Rightarrow m{x}^2-7mx+49=0$

It is a quadratic equation

Where $a=m,b=-7mandc=49$

We know that for equal roots

$D=0$

$b^2-4ac=0$

$\Rightarrow {(-7m)}^2-4\left(m\right)\left(49\right)=0$

$\Rightarrow 49m^2-49\times 4\left(m\right)=0$

$\Rightarrow 49(m^2-4m)=0$

$\Rightarrow m(m-4)=0$

Then $m=0andm=4$

Since$m=0$ is invalid value

Hence the value of $m=4$ .