Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?

share|cite|improve this question
It's not becoming, it's identifying the square of the co-efficient of 2x to be added to either side, to make the LHS a perfect square. – lab bhattacharjee Jul 29 '12 at 11:45
If you want a very intuitive, easy-to-understand derivation of the quadratic formula, you might want to check out this answer! – Shaktal Jul 29 '12 at 11:49
@Mouse Hello please tell me is there some wrong procedure in my answer?i am sure you have not downvoted,just because you have asked this question,i need you answer,you opinion – dato datuashvili Jul 29 '12 at 13:33
Sorry if I missed any comments, I haven't accessed this site for 2 days haha – Mouse Hello Aug 1 '12 at 8:51
@dato Sorry if I missed your answer, I can't seem to find it – Mouse Hello Aug 1 '12 at 8:52
up vote 7 down vote accepted

$b/2a$ does NOT become $b^2/4a^2$. All that happens in the third row is that $b^2/4a^2$ is added to both sides of the equation.

The bit about taking half of the $x$ term and squaring it is just a means of working out WHAT to add. This is often called "completing the square" - adding a constant term to an expression to turn it into a perfect square, so that one may later take its square root.

share|cite|improve this answer

Remember how to complete the square: $$Ax^2+Bx=A\left(x+\frac{B}{2A}\right)^2-\frac{B^2}{4A^2}$$

So now

$$ax^2+bx+c=0 ---- \text{complete square}$$ $$a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}=-c$$ $$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$ $$x_{1,2}+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$$ $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

share|cite|improve this answer
good answer @DonAntonio – dato datuashvili Jul 29 '12 at 13:22

$$ax^2+bx+c=0 - \text{divide by $a$ because $a\neq 0$ }$$ we get $$x^2+\frac{b}{a}x+\frac{c}{a}=0$$ $$x^2+2x\frac{b}{2a}+\frac{c}{a}=0$$ $$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}=0$$ $$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a}$$ $$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}$$ $$x^2+2x\frac{b}{2a}+\left(\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$ if in LHS we use $x=A$ and $\frac{b}{2a}=B$ then we have $$A^2+2AB+B^2=(A+B)^2$$ or $$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$ we have two values of square roote $$x_1+\frac{b}{2a}=+\sqrt{\frac{b^2-4ac}{4a^2}}$$and $$x_2+\frac{b}{2a}=-\sqrt{\frac{b^2-4ac}{4a^2}}$$ or $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

share|cite|improve this answer
Very nice Adi. Free Kosovo – Babak S. Jun 7 '13 at 5:32

From Maths is Fun.


Hope this helps

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.