How to Derive the Quadratic Formula

Co-authored by David Jia

Last Updated: December 2, 2022 References

One of the most important skills an algebra student learns is the quadratic formula, or $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.$ With the quadratic formula, solving any quadratic equation of the form $ax^{2}+bx+c=0$ becomes a simple matter of substituting the coefficients $a,b,c$ into the formula. While simply knowing the formula is often enough for many, understanding how it's derived (in other words, where it comes from) is another thing entirely. The formula is derived via "completing the square" that has other applications in math as well, so it is recommended that you be familiar with it.

Steps

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1
Start with the standard form of a general quadratic equation. While any equation with an $x^{2}$ $x^{{2}}$ term in it qualifies as quadratic, the standard form sets everything to 0. Remember that $a,b,c$ $a,b,c$ are coefficients that can be any real number, so don't substitute any numbers in for them - we want to work with the general form.^{[1]
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- $ax^{2}+bx+c=0$
- The only condition is that $a\neq 0,$ because otherwise, the equation reduces to a linear equation. See if you can find general solutions for the special cases where $b=0$ and where $c=0.$
2
Subtract $c$ from both sides. Our goal is to isolate $x.$ $x.$ To start, we move one of the coefficients to the other side, so that the left side only consists of terms with $x$ $x$ in it.^{[2]
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- $ax^{2}+bx=-c$
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3
Divide both sides by $a$ .^{[3]
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Research source} Note that we could've switched this and the previous step, and still arrived at the same place. Remember that dividing a polynomial by something means that you divide each of the individual terms. Doing so makes it easier for us to complete the square.
- $x^{2}+{\frac {b}{a}}x={\frac {-c}{a}}$
4
Complete the square. Recall that the goal is to rewrite an expression $x^{2}+2\Box x+\Box ^{2}$ $x^{{2}}+2\Box x+\Box ^{{2}}$ as $(x+\Box )^{2},$ $(x+\Box )^{{2}},$ where $\Box$ $\Box$ is any coefficient. It may not immediately be obvious to you that we can do this. To see it more clearly, rewrite ${\frac {b}{a}}x$ ${\frac {b}{a}}x$ as $2{\frac {b}{2a}}x$ $2{\frac {b}{2a}}x$ by multiplying the term by ${\frac {2}{2}}.$ ${\frac {2}{2}}.$ We can do this because multiplying by 1 does not change anything. Now we can clearly see that in our case, $\Box ={\frac {b}{2a}},$ $\Box ={\frac {b}{2a}},$ so we are only missing the $\Box ^{2}$ $\Box ^{{2}}$ term. Therefore, in order to complete the square, we add that to both sides - namely, $\left({\frac {b}{2a}}\right)^{2}={\frac {b^{2}}{4a^{2}}}.$ $\left({\frac {b}{2a}}\right)^{{2}}={\frac {b^{{2}}}{4a^{{2}}}}.$ Then, of course, we factor.^{[4]
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- ${\begin{aligned}x^{2}+2{\frac {b}{2a}}x+{\frac {b^{2}}{4a^{2}}}&={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}\\\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}\end{aligned}}$
- Here, it is clear why $a\neq 0,$ since $a$ is in the denominator, and you cannot divide by 0.
- If you need to, you can expand the left side to confirm that completing the square works.
5
Write the right side under a common denominator. Here, we want both denominators to be $4a^{2},$ $4a^{{2}},$ so multiply the ${\frac {-c}{a}}$ ${\frac {-c}{a}}$ term by ${\frac {4a}{4a}}.$ ${\frac {4a}{4a}}.$ ^{[5]
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- ${\begin{aligned}\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}}{4a^{2}}}-{\frac {4ac}{4a^{2}}}\\&={\frac {b^{2}-4ac}{4a^{2}}}\end{aligned}}$
6
Take the square root of each side. However, it is essential that you recognize that in doing so, you are actually doing two steps. When you take the square root of $d^{2},$ $d^{{2}},$ you do not get $d.$ $d.$ You actually get its absolute value, $|d|.$ $|d|.$ This absolute value is critical in getting both roots - simply putting square roots over both sides will only get you one of the roots.^{[6]
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- $\left|x+{\frac {b}{2a}}\right|={\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}$
- Now, we can get rid of the absolute value bars by putting a $\pm$ $\pm$ on the right side. We can do this because the absolute value does not distinguish between positive and negative, so they are both valid. This tidbit is why the quadratic equation allows us to get two roots.
  - $x+{\frac {b}{2a}}=\pm {\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}$
- Let's simplify this expression a bit further. Since the square root of a quotient is the quotient of the square roots, we can write the right side as ${\frac {\pm {\sqrt {b^{2}-4ac}}}{\sqrt {4a^{2}}}}.$ ${\frac {\pm {\sqrt {b^{{2}}-4ac}}}{{\sqrt {4a^{{2}}}}}}.$ Then we can take the square root of the denominator.
  - $x+{\frac {b}{2a}}={\frac {\pm {\sqrt {b^{2}-4ac}}}{2a}}$
7
Isolate $x$ by subtracting ${\frac {b}{2a}}$ from both sides.^{[7]
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- $x={\frac {-b}{2a}}\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}$
8
Write the right side under a common denominator. This nets the quadratic formula, the formula that solves any quadratic equation in standard form. This works for any $a,b,c$ $a,b,c$ and outputs an $x$ $x$ that can be real or complex. To confirm that this process works, simply follow the steps of this article in reverse order to recover standard form.^{[8]
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- $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

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What is the quadratic formula?

Donagan

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Shown in the introduction above, it's a tool used to solve a quadratic equation that is not easily solved by factoring.

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What is the cubic formula?

Donagan

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It's a formula for solving cubic equations. Take a look at math.vanderbilt.edu/schectex/courses/cubic/.

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Why is there a plus or minus in the quadratic formula?

Linnie

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When you take the square root of both sides of the equation, you need a plus or minus sign before the right side to show that the positive value squared equals (x+b/2a)² and that the negative value squared equals (x+b/2a)².The square root of x², for example, does not equal x but rather equals the absolute value of x, so the right side may be positive or negative.

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It is interesting to note that the quadratic formula holds for complex coefficients as well, though you may have to do a little more simplifying for the final answer, and the roots will no longer come in conjugate pairs. Problems with quadratic expressions are nevertheless almost always given with real coefficients.

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