Every Quadratic Equation Can Be Solved With One Formula, Even When Factorising Fails

A quadratic equation in $x$ is one whose highest power of $x$ is $2$. Its standard form is

$$a{x}^2+bx+c=0,a\ne 0$$

Because $x$ is squared, two different values can give the same result — for example, both $3$ and $-3$ satisfy $x^2=9$. The values that satisfy the equation are its roots.

This part covers standard form, factorisation, the quadratic formula and equations that reduce to quadratics.

**Expand and move every term to one side so the equation reads $a{x}^2+bx+c=0$; it is quadratic only if the coefficient of $x^2$ is not zero after simplifying.

Worked example 1.** $(x+3)(x-2)=4$

$$x^2+x-6=4x^2+x-10=0a=1,b=1,c=-10$$

Worked example 2. $2x-\frac{3}{x}=5$, with $x\ne 0$. Multiply by $x$:

$$2x^2-3=5x2x^2-5x-3=0$$

Worked example 3 — not quadratic. $(x+1{)}^2=x^2+4$

$$x^2+2x+1=x^2+42x-3=0$$

**The $x^2$ terms cancel, so this is linear.

An everyday example. A rectangular kitchen garden is $3$ m longer than it is wide and has an area of $40{\text{ m}}^2$.** With width $x$:

$$x(x+3)=40x^2+3x-40=0$$

The substance. Always simplify before deciding — an equation that looks quadratic may not be.

**Find two numbers whose product is $ac$ and whose sum is $b$, split the middle term with them, factorise by grouping, and set each factor equal to zero.

Worked example 1.** Solve $x^2+3x-40=0$.

Product $=-40$, sum $=3$: the numbers are $8$ and $-5$.

$$x^2+8x-5x-40=0x(x+8)-5(x+8)=0(x+8)(x-5)=0$$
$$x=-8\text{or}x=5$$

**For the garden, width cannot be negative, so the width is $5$ m** and the length $8$ m.

Worked example 2. Solve $6x^2-x-2=0$.

$ac=-12$, sum $=-1$: the numbers are $-4$ and $3$.

$$6x^2-4x+3x-2=02x(3x-2)+1(3x-2)=0(2x+1)(3x-2)=0$$
$$x=-\frac{1}{2}\text{or}x=\frac{2}{3}$$

An everyday example. A ball thrown straight up at a school sports day has height $h=20t-5t^2$ metres. It is $15$ m high when $5t^2-20t+15=0$, that is $(t-1)(t-3)=0$: **at $1$ s going up and $3$ s coming down.

The misconception. Never divide both sides by $x$.** From $x^2=5x$, dividing gives only $x=5$; factorising $x(x-5)=0$ also gives $x=0$.

**Substitute $a$, $b$ and $c$ into $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, work out the value under the root first, and round only at the final step.

Worked example 1.** Solve $2x^2-7x+2=0$ correct to two decimal places.

$$b^2-4ac=49-16=33\sqrt{33}=5.7446\dots$$
$$x=\frac{7+5.7446}{4}=3.1861\cdots \approx 3.19x=\frac{7-5.7446}{4}=0.3139\cdots \approx 0.31$$

To three significant figures, the roots are $3.19$ and $0.314$.

Worked example 2. Solve $x^2-4x-1=0$.

$$x=\frac{4\pm \sqrt{16+4}}{2}=\frac{4\pm 2\sqrt{5}}{2}=2\pm \sqrt{5}$$

$$x\approx 4.24\text{or}x\approx -0.24$$

An everyday example. **A carpet of area $20{\text{ m}}^2$ has length $1$ m more than twice its width.** With width $x$: $2x^2+x-20=0$.

$$x=\frac{-1+\sqrt{1+160}}{4}=\frac{-1+12.6886\dots }{4}\approx 2.92\text{ m}$$

The substance. **Rounding $\sqrt{33}$ too early** to $5.7$ gives $3.18$ instead of $3.19$ — keep extra digits until the end.

**State the values that make any denominator zero, multiply through by the lowest common denominator, simplify to $a{x}^2+bx+c=0$, solve, and reject any root that is not allowed.

Worked example 1 — reciprocal.** Solve $x+\frac{1}{x}=\frac{5}{2}$, $x\ne 0$.

$$2x^2+2=5x2x^2-5x+2=0(2x-1)(x-2)=0x=\frac{1}{2}\text{ or }2$$

Worked example 2 — fractions. Solve $\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}$, with $x\ne 0,1,2$.

Multiply by $x(x-1)(x-2)$:

$$x(x-1)+2x(x-2)=6(x-1)(x-2)$$
$$3x^2-5x=6x^2-18x+123x^2-13x+12=0$$
$$x=\frac{13\pm \sqrt{169-144}}{6}=\frac{13\pm 5}{6}x=3\text{ or }\frac{4}{3}$$

**Check $x=3$:** $\frac{1}{1}+\frac{2}{2}=2$ and $\frac{6}{3}=2$.

An everyday example. Speed and time problems, where time equals distance divided by speed, produce exactly these fraction equations.

The substance. A root that makes a denominator zero must be rejected, even if the algebra produces it.