Find constant a where quadratic equation equals zero

Question

I need to find the $a$ where $$(a-2)x^2 + (a^2 - a - 2)x + 2a^2 -4a = 0.$$ Ok, it is easy to tell that $a$ must equal 2 but... how can I find it if it's not so obvious? Do I have to take discriminant? It will be a real mess I think if I do so and I will find the $x$ I think this way...

Answer 1

$$ a^2-a-2 = (a-2)(a+1)\\\ 2a^2 - 4a = 2a(a-2) $$ the first term is trivially true

Answer 2

$$(a-2)x^2 + (a^2 - a - 2)x + 2a^2 -4a = 0=0\cdot x^2+0\cdot x+0$$

Equate the coefficients of $x^2,x$ and the constant term and solve for $a$.

You will get three equations, just involving a single variable $a$

• $ a-2=0$
• $ a^2 - a - 2=0$
• $ 2a^2 -4a=0$