The quadratic formula and the discriminant

Split­ting the mid­dle term is quick when the num­bers are friendly, but many qua­drat­ics can­not be split using whole num­bers at all. Try find­ing two inte­gers with prod­uct $-5$ and sum $-7$ for $x^2 - 7x - 5$: there are none. For every qua­dratic equa­tion, friendly or not, there is a for­mula that gives the roots directly. In this les­son we prove that for­mula, meet the dis­crim­i­nant, and use both meth­ods side by side.

The result#

The roots of $ax^2 + bx + c = 0$, where $a \neq 0$, are

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

where $\Delta = b^2 - 4ac$. The num­ber $\Delta$ (the Greek let­ter delta) is called the dis­crim­i­nant.

The sign $\pm$ means we get two roots: one using $+$ and one using $-$.

Proof by com­plet­ing the square#

The idea is to turn the left side into a per­fect square, $(x + \text{something})^2$, and then take square roots. We use one fact: if $x^2 = k$, then $x = \pm\sqrt{k}$.

Start with the equa­tion and divide every term by $a$ (allowed, since $a \neq 0$):

$$\displaystyle \begin{aligned} ax^2 + bx + c &= 0 \\ x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 \\ x^2 + \frac{b}{a}x &= -\frac{c}{a} \\ x^2 + 2(x)\left(\frac{b}{2a}\right) &= -\frac{c}{a} \end{aligned}$$

The last line writes the mid­dle term as $\displaystyle 2 \times x \times \frac{b}{2a}$, which looks like the mid­dle of $\displaystyle (x + \frac{b}{2a})^2$. To com­plete the square, add $\displaystyle \left(\frac{b}{2a}\right)^2$ to both sides:

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

Now take the square root of both sides:

$$\displaystyle \begin{aligned} x + \frac{b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm\frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{aligned}$$

That is the for­mula.

Using the for­mula#

Always do it in three steps: write the equa­tion in the form $ax^2 + bx + c = 0$, read off $a$, $b$ and $c$ with their signs, then sub­sti­tute.

Exam­ple 1#

Find the roots of $x^2 - 3x + 2 = 0$.

Here $a = 1$, $b = -3$, $c = 2$.

$$\displaystyle \begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{3 \pm \sqrt{9 - 4(1)(2)}}{2(1)} \\ &= \frac{3 \pm \sqrt{1}}{2} = \frac{3 \pm 1}{2} \end{aligned}$$

So $\displaystyle x = \frac{3 - 1}{2} = 1$ or $\displaystyle x = \frac{3 + 1}{2} = 2$.

Check by split­ting the mid­dle term: $x^2 - x - 2x + 2 = x(x - 1) - 2(x - 1) = (x - 1)(x - 2)$, which gives the same roots $1$ and $2$.

Notice that $-b = -(-3) = 3$. Get­ting the sign of $-b$ wrong is the most com­mon error with this for­mula.

Exam­ple 2#

Find the roots of $x^2 - 7x - 5 = 0$.

Here $a = 1$, $b = -7$, $c = -5$.

$$\displaystyle \begin{aligned} x &= \frac{7 \pm \sqrt{49 - 4(1)(-5)}}{2(1)} \\ &= \frac{7 \pm \sqrt{49 + 20}}{2} \\ &= \frac{7 \pm \sqrt{69}}{2} \end{aligned}$$

So the roots are $\displaystyle \frac{7 - \sqrt{69}}{2}$ and $\displaystyle \frac{7 + \sqrt{69}}{2}$. Since $69$ is not a per­fect square, the roots are irra­tional, which is exactly why split­ting the mid­dle term could not work.

The dis­crim­i­nant#

The num­ber under the root, $\Delta = b^2 - 4ac$, tells us what kind of roots to expect before we fin­ish the work­ing. For an equa­tion whose coef­fi­cients $a$, $b$, $c$ are inte­gers:

• If $\Delta$ is a per­fect square (as in Exam­ple 1, $\Delta = 1$), the roots are ratio­nal and the qua­dratic can be split using ratio­nal num­bers.
• If $\Delta$ is pos­i­tive but not a per­fect square (as in Exam­ple 2, $\Delta = 69$), the roots are irra­tional.
• If $\Delta$ is neg­a­tive, there is no real square root, so the equa­tion has no real roots. You will study this case fully with qua­dratic equa­tions.

Split­ting with a lead­ing coef­fi­cient#

When $a$ is not $1$, split­ting still works: look for two num­bers whose prod­uct is $a \times c$ and whose sum is $b$.

Exam­ple 3#

Solve $6x^2 - 13x + 6 = 0$.

Prod­uct $6 \times 6 = 36$, sum $-13$. The fac­tor pairs of $36$ are $1 \times 36$, $2 \times 18$, $3 \times 12$, $4 \times 9$, and $4 + 9 = 13$, so we use $-4$ and $-9$.

$$\begin{aligned} 6x^2 - 4x - 9x + 6 &= 0 \\ 2x(3x - 2) - 3(3x - 2) &= 0 \\ (3x - 2)(2x - 3) &= 0 \end{aligned}$$

So $\displaystyle x = \frac{2}{3}$ or $\displaystyle x = \frac{3}{2}$.

By the for­mula, $\Delta = 169 - 144 = 25$ and $\displaystyle x = \frac{13 \pm 5}{12}$, which gives $\displaystyle \frac{18}{12} = \frac{3}{2}$ and $\displaystyle \frac{8}{12} = \frac{2}{3}$. Both meth­ods agree.

Exam­ple 4#

Solve $x^2 - 5x - 24 = 0$.

Prod­uct $-24$, sum $-5$. A neg­a­tive prod­uct means one num­ber is neg­a­tive: $-8 \times 3 = -24$ and $-8 + 3 = -5$.

$$\begin{aligned} x^2 - 8x + 3x - 24 &= 0 \\ x(x - 8) + 3(x - 8) &= 0 \\ (x + 3)(x - 8) &= 0 \end{aligned}$$

So $x = -3$ or $x = 8$.

Prac­tice#

1. Find the dis­crim­i­nant of $2x^2 - 7x + 3$.
2. Solve $x^2 - 6x + 5 = 0$ using the for­mula.
3. Solve $2x^2 - x - 6 = 0$ using the for­mula.
4. Solve $2x^2 + 7x + 3 = 0$ using the for­mula.
5. Solve $x^2 + 4x - 1 = 0$.
6. Solve $3x^2 + 5x - 2 = 0$ by split­ting the mid­dle term, and check with the for­mula.
7. Which of the fol­low­ing are qua­dratic equa­tions? (i) $2x^2 - 5 = 0$ (ii) $\displaystyle x^2 - \frac{1}{x^2} = 5$ (iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
8. If $\displaystyle x = -\frac{1}{2}$ is a zero of $p(x) = 8x^3 - ax^2 - x + 2$, find $a$.
9. Solve $44x^2 + 65x + 24 = 0$.

Answers#