Solve the equation :- $[4ax - (a + b)](a + b) = 0$ , where $a$ and $b$ are constants .
What I Tried:- An usual idea will be to break up the expression. So I get:-
$$[4ax - (a + b)](a + b) = 0$$ $$\implies 4ax(a + b) - (a + b)^2 = 0$$ $$\implies 4a^2x + 4abx - a^2 + 2ab - b^2 = 0$$
From, here I don't know how to proceed. Can anyone help /
If $a+b=0$, any $x$ is a solution.
If $a+b\ne 0$, divide both sides by $a+b$. You get: $4ax-(a+b)=0$, i.e. $4ax=a+b$.
If $a=0$, then you get $0=a+b$. As we are already restricted to the case $a+b\ne 0$, this is impossible.
If $a\ne 0$, then $x=\frac{a+b}{4a}=\frac14+\frac{b}{4a}$.
Altogether: if $a+b=0$, every $x$ is a solution. If $a+b\ne 0$ and $a=0$, then the equation has no solutions. If $a+b\ne 0$ and $a\ne 0$, then the unique solution is $x=\frac14+\frac{b}{4a}$.