How does the DISCRIMINANT really work?

Question

I have the following equation (depended on param a)

$(a^2-2a)x^2+2ax-1=0$

I want to find out what the behavior of this equation after changing the value (by behavior I mean finding out how will the root count change when I put x or y instead of a). So for that purpose I thought it might be useful to find the D. If i'm not wrong when we have b = 2k ,then we can simply use this formula:

$D = k^2-ac = a^2+(a^2-2a)=2a^2-2a=2a(a-1)$

so after having this done, I thought maybe I can find out where my equation has only 1 root:

$2a(a-1)=0, a=0, 1$

but after putting for example a = 0, I quickly noticed that there my equation has no roots(real roots I think).

$a=0, -1 =0\Rightarrow x\in\varnothing$

Can someone explain me what I've missed here? Shouldn't a = 0, work just fine (I mean give me only 1 root)?

Answer 1

The complete discussion of the equation $Ax^2+Bx+C=0$ in the real variable $x$ and in terms of the real parameters $A,B,C$ is:

• if $A\ne 0$ and $B^2-4AC\ge0$, then the solution is $x=\frac{-B-\sqrt{B^2-4AC}}{2A}\lor x=\frac{-B+\sqrt{B^2-4AC}}{2A}$ (and these are equal if and only if $B^2-4AC=0$).

• if $A\ne 0$ and $B^2-4AC<0$, then there are no solutions in $\Bbb R$.

• if $A=0$ and $B\ne 0$, then the solution is $x=-\frac CB$.

• if $A=0$ and $B=0$ and $C\ne 0$, then there are no solutions

• if $A=B=C=0$, then all $x\in \Bbb R$ solve the equation.

So it isn't quite a matter of discriminant, but rather of what happens to the reduced equation when the leading term vanishes.

Remark: You are correct in saying that when $B$ is in the form $B=2\beta$, it's easier to calculate directly $\frac\Delta4=\left(\frac B2\right)^2-AC=\beta^2-AC$ and use $\frac{-B\pm\sqrt{\Delta}}{2A}=\frac{-\beta\pm\sqrt{\Delta/4}}{A}$.

Answer 2

The discriminant is valid only for quadratic equations. If $a=0$ then in your case the equation constant(degree 0). In this kind of exercise you should always impose the quadratic coefficient(that in your case is $a^2-2a$) to be non-zero, and study separately the zero case.(A little precisation the one you calculated isn't actually $D$, but $\frac D4$, the trick works because dividing by $4$ doesn't change the sign of the discriminant that it's actually the only thing that matters in this context, clearly if you choose to use $\frac D4$ the quadratic formula must be slightly modified)

Answer 3

Let the equation be $Ax^2 +Bx +C=0$

The discriminant $(D)$ you have calculated is wrong.

$D = B^2-4AC= (2a)^2 - 4(-1)(a^2 - 2a) $

$D = 4a^2+4a^2-8a = 8a(a-1)$

Secondly, $D=0$ implies 1 root only if $A ≠ 0 $
because the roots are $\frac{-B+\sqrt{D}}{2A},\frac{-B+\sqrt{D}}{2A}$

The denominator cannot be $0$.

Therefore, to make the right use of determinant $A ≠ 0 $