Quadratic formula

After factorisation (when it works) and completing the square (which always works but is slow), the quadratic formula is the practical workhorse for solving quadratics.

The formula

For $a x^2 + b x + c = 0$ with $a \ne 0$, $\boxed{x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}.}$

The number $D = b^2 - 4 a c$ under the square root is called the discriminant.

How to use

1. Bring the equation into standard form $a x^2 + b x + c = 0$.
2. Identify $a, b, c$ , be careful with signs.
3. Compute the discriminant $D = b^2 - 4 a c$.
4. If $D < 0$: no real roots, stop.
5. If $D \ge 0$: roots are $x = \dfrac{-b + \sqrt{D}}{2 a}$ and $x = \dfrac{-b - \sqrt{D}}{2 a}$.

A few tips:

• Always double-check signs of $b$ and $c$ before computing the discriminant.
• If the equation has decimal or fractional coefficients, multiply through to clear them , the formula behaves better with integers.
• If the radical $\sqrt{D}$ is not a perfect square, leave it in surd form unless the problem asks for a decimal.

Worked examples

Example 1. Solve $2 x^2 - 5 x + 3 = 0$.

$a = 2, b = -5, c = 3$. $D = 25 - 24 = 1$.

$x = \dfrac{5 \pm 1}{4} = 3/2, 1$.

Example 2. Solve $3 x^2 - 6 x + 2 = 0$.

$a = 3, b = -6, c = 2$. $D = 36 - 24 = 12$. $\sqrt{D} = 2 \sqrt{3}$.

$x = \dfrac{6 \pm 2\sqrt{3}}{6} = 1 \pm \dfrac{\sqrt{3}}{3}$, i.e., $x = 1 \pm \dfrac{1}{\sqrt{3}}$.

Example 3. Solve $x^2 + x + 1 = 0$.

$D = 1 - 4 = -3 < 0$. No real roots.

Example 4. Solve $4 x^2 + 4 \sqrt{3} x + 3 = 0$.

$a = 4, b = 4 \sqrt{3}, c = 3$. $D = 48 - 48 = 0$. Repeated root $x = -b/(2 a) = -4 \sqrt{3}/8 = -\sqrt{3}/2$.

Example 5. Solve $\sqrt{3} x^2 + 10 x + 7 \sqrt{3} = 0$.

$a = \sqrt{3}, b = 10, c = 7 \sqrt{3}$. $D = 100 - 4 \sqrt{3} \cdot 7 \sqrt{3} = 100 - 84 = 16$. $\sqrt{D} = 4$.

$x = \dfrac{-10 \pm 4}{2 \sqrt{3}} = \dfrac{-6}{2 \sqrt{3}}, \dfrac{-14}{2 \sqrt{3}} = -\sqrt{3}, -\dfrac{7 \sqrt{3}}{3} = -\sqrt{3}, -\dfrac{7}{\sqrt{3}}$.

(Rationalising, the second root is $-\dfrac{7 \sqrt{3}}{3}$.)

Try it yourself

1. Solve: $x^2 - 7 x + 12 = 0$ using the formula.
2. Solve: $x^2 - 4 x + 1 = 0$.
3. Solve: $4 x^2 - 4 a^2 x + (a^4 - b^4) = 0$.
4. Solve: $x^2 + 2 \sqrt{2} x - 6 = 0$.
5. Solve: $2 x^2 + x - 4 = 0$.
6. Solve: $5 x^2 - 3 x - 2 = 0$.
7. Solve: $3 x^2 + 2 \sqrt{5} x - 5 = 0$.
8. For $a x^2 + b x + c = 0$, what does $D > 0$ tell you? $D = 0$? $D < 0$?
9. Solve: $\dfrac{1}{x - 1} - \dfrac{1}{x + 5} = \dfrac{6}{7}$.
10. A train, travelling at a uniform speed for $360$ km, would have taken $48$ minutes less to travel the same distance if its speed were $5$ km/h more. Set up a quadratic and solve.

Pitfalls / Insight

• Sign of $b$. A typical mistake is plugging $b = 5$ instead of $b = -5$ when the equation has $-5 x$.
• Discriminant under the square root , never take the root before computing the discriminant carefully.
• Surds and rationalisation , write the final answer in simplest form. Convert $\dfrac{1}{\sqrt{3}}$ to $\dfrac{\sqrt{3}}{3}$ if asked.

Insight. The quadratic formula is the tool. If you can't see a clean factorisation in $15$ seconds, switch to the formula , and let the discriminant tell you what kind of answer to expect.