Standard form and factorisation

Definitions

A quadratic equation in $x$ is any equation that can be rewritten in the standard form $a x^2 + b x + c = 0,$ where $a, b, c$ are real, $a \ne 0$. The numbers $a, b, c$ are the coefficients; $a$ is the leading coefficient, $b$ the coefficient of $x$, and $c$ the constant term.

A root (or solution) of the equation is a number $\alpha$ such that $a \alpha^2 + b \alpha + c = 0$. Equivalently, it is a zero of the polynomial $a x^2 + b x + c$.

A quadratic equation has at most two roots, by what we proved in Chapter 2.

Strategy: factorisation by splitting the middle term

The fastest method when the quadratic has rational roots is to factor the left side into two linear factors. The key technique is splitting the middle term.

To factor $a x^2 + b x + c$:

1. Find two numbers $p$ and $q$ such that $p + q = b$ and $p q = a c$.
2. Rewrite $b x$ as $p x + q x$: $a x^2 + p x + q x + c.$
3. Group: $(a x^2 + p x) + (q x + c) = x (a x + p) + \frac{q}{?}\cdots,$ and factor each group. (You may need to scale: typically the factor in both groups comes out the same.)
4. You get $a x^2 + b x + c = (a_1 x + b_1)(a_2 x + b_2)$.
5. Set each linear factor to zero to find the roots.

The hunt for the two numbers $p, q$ is just a quick mental arithmetic exercise: find a factor pair of $a c$ that sums to $b$.

For example, to factor $3 x^2 + 7 x + 2$, we look for $p, q$ with $p q = 6$ and $p + q = 7$. The pair is $(1, 6)$. So $3 x^2 + x + 6 x + 2 = x(3 x + 1) + 2(3 x + 1) = (3 x + 1)(x + 2).$ Roots: $x = -1/3, -2$.

When $a = 1$ this collapses to the simpler problem of finding two numbers whose product is $c$ and sum is $b$ , the classical "split the middle term".

If the quadratic does not factor over the rationals (no clean integer/rational roots), this method is not enough; you'll need completing the square or the formula, coming up next.

Worked examples

Example 1. Solve $x^2 - 5 x + 6 = 0$.

Need two numbers with product $6$, sum $-5$: $(-2, -3)$.

$x^2 - 2 x - 3 x + 6 = x(x - 2) - 3(x - 2) = (x - 2)(x - 3)$. Roots: $x = 2, 3$.

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

Need $p q = -6, p + q = 5$. Try $(6, -1)$: $6 \cdot (-1) = -6$, $6 + (-1) = 5$. ✓

$2 x^2 + 6 x - x - 3 = 2 x(x + 3) - (x + 3) = (x + 3)(2 x - 1)$. Roots: $x = -3, 1/2$.

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

$p q = -12, p + q = -1$. Try $(3, -4)$: $-12, -1$. ✓

$6 x^2 + 3 x - 4 x - 2 = 3 x(2 x + 1) - 2(2 x + 1) = (2 x + 1)(3 x - 2)$. Roots: $x = -1/2, 2/3$.

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

This is $(2 x)^2 + 2 \cdot 2 x \cdot \sqrt{3} + (\sqrt{3})^2 = (2 x + \sqrt{3})^2$. So $(2 x + \sqrt{3})^2 = 0$, giving the (repeated) root $x = -\sqrt{3}/2$.

Example 5. Solve $\dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30}$.

Combine: $\dfrac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \dfrac{11}{30} \Rightarrow \dfrac{-11}{(x + 4)(x - 7)} = \dfrac{11}{30}$.

Cross-multiply: $-30 = 11(x + 4)(x - 7) / 11 \cdot (\text{rearrange}) \Rightarrow (x + 4)(x - 7) = -30$. Expand: $x^2 - 3 x - 28 = -30 \Rightarrow x^2 - 3 x + 2 = 0 \Rightarrow (x - 1)(x - 2) = 0$. Roots $x = 1, 2$.

Try it yourself

1. Solve: $x^2 + 11 x + 30 = 0$.
2. Solve: $2 x^2 - 7 x + 3 = 0$.
3. Solve: $3 x^2 - 14 x - 5 = 0$.
4. Solve: $x^2 - (\sqrt{3} + 1) x + \sqrt{3} = 0$.
5. Solve: $x^2 + 6 x + 9 = 0$.
6. Solve: $5 x^2 - 6 x - 2 = 0$ (use factorisation if possible; else state).
7. Solve: $\dfrac{x - 3}{x + 3} + \dfrac{x + 3}{x - 3} = \dfrac{10}{3}$.
8. Find $k$ for which $k x^2 - 7 x + 2 = 0$ has $x = 2$ as a root.
9. Solve: $x^2 - 2 \sqrt{2} x - 6 = 0$ (try to factor).
10. Solve: $(x - 3)(x + 7) = 0$. (Already factored.)

Pitfalls / Insight

• Always check the signs of $p$ and $q$ , getting them wrong gives the wrong factorisation.
• Watch for repeated roots. $4 x^2 + 4 \sqrt{3} x + 3$ has just one root with multiplicity $2$.
• Verify by substitution before declaring final answers.

Insight. Factorisation works only when the quadratic has nice roots. If you spend more than a minute hunting for $p, q$, the equation probably doesn't factor cleanly , switch to the formula.