So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me


$ax^2+bx+c=0 (a\neq0)$






$\therefore x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$


Modern texts tell you how to find the roots by completing the square. It makes a nice proof but a terrible derivation. This is how the quadratic equation is found from a more reasoned and historical point of view.

Multiplying $f(x)$ by a constant does not change the location of the roots. So feel free to divide $ax^2 + bx + c$ through by $a$ and forget that you did it.

You should know how to shift equations to the left and right. $f(x - k)$ is $f$ shifted to the right $k$ amount.

You should know the sum of the roots of any polynomial just by looking at it. For a polynomial with the first coefficient being 1, The second coefficient (in this case b) is the (negative) sum of the roots, the last coefficient is the product.

Shift the polynomial to the right so that the roots are equidistant from the y axis. How to do this? Shift the polynomial left by the average of the roots, or $-b/2$. The is the same as finding $g(x) = f(x - b/2)$. Now it is easy to find the roots of $g(x)$. Then just remember to shift the roots back to the right by substracting $b/2$ from your final answer.

  • $\begingroup$ Completing the square is how Al-Khwarizmi did it. Except he was completing a real geometric square. $\endgroup$ – André Nicolas Oct 5 '13 at 19:01
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    $\begingroup$ this is not clearer, easier to follow or more enlightening than completing the square. $\endgroup$ – Jonathan Oct 5 '13 at 23:28
  • $\begingroup$ @Jonathan: I'd also add that the shifting by -b/2 is effectively the same as is done in completing the square... $\endgroup$ – Chris Oct 6 '13 at 0:36
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    $\begingroup$ Actually it is clearer if you already know basic facts about polynomials, like sums and shifting techniques. Secondly, it is enlightening because cubics and quartics solving start with the same argument, shifting the polynomial to eliminate the 2nd coefficient. You can't complete the square to solve a cubic. ...but obviously what is clear and insightful differs from person to person... $\endgroup$ – DanielV Oct 6 '13 at 6:57

For a lot of ways to solve quadratic equations, see Pat Ballew's manuscript Solving Quadratic Equations.

Here's a way to get the quadratic formula if you partially remember it. (I don't know off-hand if this is one of the methods in Pat Ballew's list.)

We remember that the quadratic formula gives us results that can be put into the form $A \pm \sqrt{B}.$ Note that we don't need to incorporate a constant factor with the square root, since we can bring the square of any such factor inside the square root: $7\sqrt{3}$ is equal to $\sqrt{7^2 \cdot 3}.$

Thus, we know the answer will be

$$x \; = \; A \pm \sqrt{B}$$

Isolating the radical and squaring gives:

$$(x - A)^2 \; = \; \left( \pm \sqrt{B} \right)^2$$

$$(x - A)^2 \; = \; B$$

$$(x - A)^2 - B \; = \; 0$$

$$ x^2 - 2Ax + (A^2 - B) \; = \; 0 $$

Now let's solve $x^2 + bx + c = 0.$ [By dividing through by the coefficient of $x^2,$ we may assume the coefficient of $x^2$ is $1.]$

Comparing the coefficients of $x^2 + bx + c = 0$ with the coefficients of the last displayed equation, we see that $x = A \pm \sqrt{B},$ where $-2A = b$ and $(A^2 - B) = c.$ Solving for $A$ and $B,$ we get $A = -\frac{b}{2}$ and $B = \frac{b^2}{4} - c,$ and now we have the solution for $x$ in terms of the equation's coefficients $b$ and $c.$

  • $\begingroup$ This is a truly elegant way to describe it! $\endgroup$ – String Oct 5 '13 at 19:12
  • $\begingroup$ @String: That looks really clumsy to me compared to completing the square method described by "눉송이". $\endgroup$ – Chris Oct 6 '13 at 0:35
  • $\begingroup$ @Chris: It is just due to differences in taste. IMO it is always preferable to communicate reasoning rather than to show (tedious) workings. $\endgroup$ – String Oct 6 '13 at 0:42
  • $\begingroup$ @String: This is definitely an agree to disagree moment. :) I think the other answer shows reasoning much better. Especially since this one requires you to remember the form of the answer as well. :) But yes, as you say clearly a subjective opinion and I didn't mean to suggest otherwise in my first comment. :) $\endgroup$ – Chris Oct 6 '13 at 0:49
  • $\begingroup$ @Chris: Thanks for the comments! I did not regard your first comment any different to what you just confirmed :) $\endgroup$ – String Oct 6 '13 at 0:56

$ax^2+bx+c = a(x^2+\frac{bx}{a}+\frac{c}{a})=a((x+\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2})$

Equate this to zero

NOTE: $a\neq0, $ if it is then the equation will not be quadratic.

you get $(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}$

which gives the desired $$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$


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