# Proving Quadratic Formula

purplemath.com 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 $\dfrac{b^2}{4a^2}$?

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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

## 5 Answers

$b/2a$ 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.

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

So now

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

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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 $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

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Very nice Adi. Free Kosovo – Babak S. Jun 7 '13 at 5:32

From Maths is Fun.

Hope this helps

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