How does the DISCRIMINANT really work? I have the following equation (depended on param a) P

Slade Higgins

Slade Higgins

Answered question

2022-04-22

How does the DISCRIMINANT really work?
I have the following equation (depended on param a)
(a22a)x2+2ax1=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=k2ac=a2+(a22a)=2a22a=2a(a1)
so after having this done, I thought maybe I can find out where my equation has only 1 root:
2a(a1)=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=0x
Can someone explain me what I've missed here? Shouldn't a=0, work just fine (I mean give me only 1 root)?

Answer & Explanation

Killian Curry

Killian Curry

Beginner2022-04-23Added 18 answers

The complete discussion of the equation Ax2+Bx+C=0 in the real variable x and in terms of the real parameters A,B,C is:
- if A0 and B24AC0, then the solution is x=BB24AC2Al or x=B+B24AC2A (and these are equal if and only if B24AC=0).
- if A0 and B24AC<0, then there are no solutions in R.
- if A=0 and B0, then the solution is x=CB.
- if A=0 and B=0 and C0, then there are no solutions
- if A=B=C=0, then all xR 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.

belamontern9i

belamontern9i

Beginner2022-04-24Added 19 answers

Step 1
Let the equation be Ax2+Bx+C=0
The discriminant (D) you have calculated is wrong.
D=B24AC=(2a)24(1)(a22a)
D=4a2+4a28a=8a(a1)
Step 2
Secondly, D=0 implies 1 root only if A0 because the roots are B+D2A,B+D2A
The denominator cannot be 0.
Therefore, to make the right use of determinant A0.

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