Geometric Proof of the Quadratic Formula

The quadratic formula allows us to easily obtain the roots of any quadratic equation $ax^2 + bx + c =0$. There is a very nice proof of this formula that uses geometry to give an intuitive understanding of this result. We show this proof below alongside the typical purely algebraic proof.

Theorem 1 (The Quadratic Formula): Let $a$, $b$, and $c$ be real numbers with $a \neq 0$. Then the solutions to the quadratic equation $ax^2 + bx + c = 0$ are $\displaystyle{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$.

We require that $a \neq 0$ to ensure that $ax^2 + bx + c = 0$ is indeed a quadratic equation. If $a = 0$ then $ax^2 + bx + c = 0$ is the same as $bx + c = 0$, which is a linear equation.

• Proof 1 (Geometric): Consider the quadratic equation $ax^2 + bx + c = 0$ with $a$, $b$, and $c$ as real numbers with $a \neq 0$. Then:

(1)

\begin{align} \quad ax^2 + bx + c &= 0 \\ \quad ax^2 + bx &= -c \\ \quad x^2 + \frac{b}{a}x &= -\frac{c}{a} \end{align}

• We illustrate the equation above as follows. We denote $x^2$ to be the area of square. Therefore, this square has side length $x$.

• We denote $\displaystyle{\frac{b}{a}x}$ to be the area of a rectangle whose side lengths are $x$ and $\displaystyle{\frac{b}{a}}$.

• Lastly, we denote $\displaystyle{-\frac{c}{a}}$ to denote the area of a rectangle whose side lengths are unimportant.

• Consider the rectangle $\displaystyle{\frac{b}{a}x}$. We separate this rectangle into two equal rectangles whose side lengths are $\displaystyle{\frac{b}{2a}}$ and $x$:

• We take the first of these rectangles and attach the side with side length $x$ to the top of the square $x^2$. WE take the second of these rectangles and attach the side with side length $x$ to the right side of the square $x^2$. We add a smaller square with side length $\displaystyle{\frac{b}{2a}}$ (and area $\displaystyle{\frac{b^2}{4a^2}}$ to the lefthand side of the equation and we add the same amount of area to the righthand side of the equation to balance the equality:

• From above we get the following equation:

(2)

\begin{align} \quad \left ( x + \frac{b}{2a} \right )^2 &= -\frac{c}{a} + \frac{b^2}{4a^2} \\ &= - \frac{-4ac}{4a^2} + \frac{b^2}{4a^2} \\ &= \frac{b^2 - 4ac}{4a^2} \end{align}

• Therefore:

(3)

\begin{align} \quad x + \frac{b}{2a} &= \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} \\ x &= - \frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \blacksquare \end{align}

• Proof 2 (Algebraic): Consider the quadratic equation $ax^2 + bx + c = 0$ with $a$, $b$, and $c$ as real numbers with $a \neq 0$. Then:

(4)

\begin{align} \quad ax^2 + bx + c &= 0 \\ ax^2 + bx &= -c \\ x^2 + \frac{b}{a}x &= -\frac{c}{a} \\ x^2 + \frac{b}{a}x + \left ( \frac{b}{2a} \right )^2 &= -\frac{c}{a} + \left ( \frac{b}{2a} \right )^2 \\ x^2 + \frac{b}{2a}x + \frac{b}{2a}x + \left ( \frac{b}{2a} \right )^2 &= \frac{b^2 - 4ac}{4a} \\ x \left ( x + \frac{b}{2a} \right ) + \frac{b}{2a} \left ( x + \frac{b}{2a} \right ) &= \frac{b^2 - 4ac}{4a} \\ \left ( x + \frac{b}{2a} \right )^2 &= \frac{b^2 - 4ac}{4a} \\ x + \frac{b}{2a} &= \pm \sqrt{\frac{b^2 - 4ac}{4a}} \\ x &= -\frac{b}{2a} + \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \blacksquare \end{align}