The quadratic formula

One method for every equation

Completing the square works, but the steps are the same each time. So let us do them once, with letters, for the general equation

Deriving the formula

1. Divide by $a$. $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$.
2. Move the number term. $x^2 + \dfrac{b}{a}x = -\dfrac{c}{a}$.
3. Add the square of half the coefficient of $x$. Half of $\dfrac{b}{a}$ is $\dfrac{b}{2a}$. Add $\dfrac{b^2}{4a^2}$ to both sides.
4. Write the square.
   • $\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2}{4a^2} - \dfrac{c}{a}$
   • $= \dfrac{b^2 - 4ac}{4a^2}$
5. Take the square root. If $b^2 - 4ac \geq 0$, then $x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}$.
6. Solve for $x$. This is the quadratic formula:

The plus sign gives one root and the minus sign gives the other.

Using the formula

Solve $3x^2 + 2x - 8 = 0$.

1. Read off the numbers. $a = 3$, $b = 2$, $c = -8$. Keep the signs.
2. Work out $b^2 - 4ac$. $4 - 4 \times 3 \times (-8) = 4 + 96 = 100$.
3. Put them in. $x = \dfrac{-2 \pm 10}{6}$.
4. The two roots. $x = \dfrac{8}{6} = \dfrac{4}{3}$, or $x = \dfrac{-12}{6} = -2$.

Check. $3 \times 4 - 4 - 8 = 0$ for $x = -2$. It works.

Now one that does not factorise: $x^2 - 4x + 1 = 0$. Here $b^2 - 4ac = 16 - 4 = 12$, so

the same answer that completing the square gave in the last lesson.

A problem that needs the formula

A ball is thrown upwards. After $t$ seconds its height is $h = 20t - 5t^2$ metres. When is it $12\ \text{m}$ above the ground?

1. Write the equation. $20t - 5t^2 = 12$, so $5t^2 - 20t + 12 = 0$.
2. The discriminant. $400 - 4 \times 5 \times 12 = 400 - 240 = 160$.
3. The formula. $t = \dfrac{20 \pm \sqrt{160}}{10} = 2 \pm \dfrac{\sqrt{160}}{10}$.
4. The numbers. $\sqrt{160} \approx 12.65$, so $t \approx 3.26\ \text{s}$ or $t \approx 0.74\ \text{s}$.

Both answers make sense. The ball passes $12\ \text{m}$ on the way up, after about $0.74\ \text{s}$. It passes it again on the way down, after about $3.26\ \text{s}$.

Practise with new numbers

Try it in Python

The function solve uses the formula. Then the program puts each root back into the equation. The left side should come out as $0$, or as a tiny number such as $10^{-16}$ from rounding.

Common mistakes

• Losing a sign. With $b = -4$, $-b = 4$. And $b^2 = 16$, never $-16$.
• Dividing only the root by $2a$. The whole top, $-b \pm \sqrt{b^2 - 4ac}$, is divided by $2a$.
• Forgetting that $c$ can be negative. In $3x^2 + 2x - 8 = 0$, $c = -8$, so $-4ac$ is $+96$.
• Writing $a$ when it is $1$ and hidden. In $x^2 - 4x + 1 = 0$, $a = 1$.

Remember these

• For $ax^2 + bx + c = 0$: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Read $a$, $b$ and $c$ with their signs, work out $b^2 - 4ac$ first, then the two roots.
• Check each root by putting it back, and check that it makes sense in the problem.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In $2x^2 - 7x + 3 = 0$, what are $a$, $b$ and $c$?

2. Solve $x^2 + 2x - 4 = 0$.

3. Why does the formula have a $\pm$ sign?

4. Solve $5x^2 - 2x - 3 = 0$.

5. Ishaan writes $x = -b \pm \dfrac{\sqrt{b^2 - 4ac}}{2a}$. What is wrong?