Maybe the question is very trivial in a sense. So, it doesn't work for anyone. A few years ago, when I was a seventh-grade student, I had found a quadratic formula for myself. Unfortunately, I didn't have the chance to show it to my teacher at that time and later I saw that it was "trivial". I saw this formula again by chance while mixing my old notebooks. I wonder if this simple formula is used somewhere.
Let's remember the original method first:
ax2+bx+c=0, a≠04a2x2+4abx+4ac=04a2x2+4abx=−4ac4a2x2+4abx+b2=b2−4ac(2ax+b)2=b2−4ac2ax+b=±√b2−4acx1,2=±√b2−4ac−b2ax1,2=−b±√b2−4ac2aax2+bx+c=0, a≠04a2x2+4abx+4ac=04a2x2+4abx=−4ac4a2x2+4abx+b2=b2−4ac(2ax+b)2=b2−4ac2ax+b=±b2−4ac−−−−−−−√x1,2=±b2−4ac−−−−−−−√−b2ax1,2=−b±b2−4ac−−−−−−−√2a
In fact, the "meat" of this method is as follows:
ax2+bx+c=0, a≠0x2+bax+ca=0(x+b2a)2−(b2a)2+ca=0(x+b2a)2=b24a2−ca(x+b2a)2=b2−4ac4a2x+b2a=±√b2−4ac2ax1,2=−b±√b2−4ac2a
Now, we know that if one of the roots for ax2+bx+c=0 is x=0, then our equation is equivalent to ax2+bx=0. No special formula is required to solve the last equation.
In this sense, I am setting off by accepting that x≠0.
ax2+bx+c=0, a≠0a+bx+cx2=0cx2+bx+a=04c2x2+4bcx+4ac=04c2x2+4bcx=−4ac4c2x2+4bcx+b2=b2−4ac(2cx+b)2=b2−4ac2cx+b=±√b2−4ac2cx=−b±√b2−4acx1,2=2c−b±√b2−4ac
Let's rewrite the well-known general formula as follows:
−b±√b2−4ac2a=−b∓√b2−4ac2a
If we accept c≠0, then we have:
2c−b±√b2−4ac=−b∓√b2−4ac2a⟺(−b±√b2−4ac)×(−b∓√b2−4ac)=4ac⟺−(b∓√b2−4ac)×(−(b±√b2−4ac))=4ac⟺(b∓√b2−4ac)×(b±√b2−4ac)=4ac⟺b2−(b2−4ac)=4ac⟺4ac=4ac.
Since we have accepted x≠0 before, this formula cannot work completely for c=0.
If c=0, then we have:
x1=0−2b=0 which imply, one of the roots is correct.
x2=00=undefined which imply, the second root is incorrect.
These are interesting points for an untutored person like me. On the other hand, they are trivial.
If the Δ (Discriminant) is zero, then there is exactly one real root, sometimes called a repeated or double root.
Δ=b2−4ac or D=b2−4ac and D=0, then we have :
From the formula x1,2=−b±√D2a,
x=x1=x2=−b2a=−b2a
From the formula x1,2=2c−b±√D,
x=x1=x2=−2cb=−2cb
which both are equal.
x=x1=x2=−b2a=−2cb⟹b2=4ac⟹b2−4ac=0.
The original formula does not work for a=0. However, the alternative formula also works when a=0. The important point is that we should be careful not to make the denominator zero. In other words,
If a=0 and b>0 then we write:
x=2c−b−√b2=−cb
If a=0 and b<0 then we write:
x=2c−b+√b2=−cb
Maybe in some special cases, can this formula be more useful than its own alternative? (I assume the formula I found here is correct.)