2.14 - Solving linear and quadratic equations

Choose a useful method for a linear or quadratic equation, keep exact answers where they are needed and check which solutions fit a problem. You will connect factorisation, completing the square, the quadratic formula and graphical solutions, rather than treating them as unrelated tricks.

Keeping an equation balanced

A rectangle has area $40\,\mathrm{cm}^2$ , and its length is $3\,\mathrm{cm}$ greater than its width. Why will the algebra produce two possible widths when only one rectangle is possible? We will return to this question after separating two kinds of equation.

An equation says that two expressions have equal values. A solution is a value that makes that statement true. Solving changes the equation into a simpler equivalent statement, while keeping exactly the same solutions.

You already know

Expand brackets carefully, collect like terms and calculate with exact fractions. Doing the same addition or subtraction to both sides, or multiplying or dividing both sides by the same non-zero number, preserves equality.

A linear equation in one unknown simplifies to $ax+b=0$ , with $a\ne0$ . Its unknown has power $1$ . For example, $3(x+2)=2x+11$ is linear; $x^2+3x=4$ is quadratic. Collecting terms can also reveal a boundary case: $2x+1=2x+1$ is true for every real $x$ , whereas $2x+1=2x+3$ has no solution. Neither simplifies to a linear equation with a non-zero coefficient of $x$ .

Clear numerical denominators before collecting terms. Multiplication must reach every term on each side.

Worked example 1: clearing a denominator

Question. Solve $\dfrac{3x+2}{4}=x-1$ .

Plan. Multiply both sides by $4$ , then collect the unknown terms on one side.

\begin{aligned} 3x+2&=4x-4 && \text{multiply both sides by }4 \\ 6&=x && \text{subtract }3x\text{ and add }4 \end{aligned}

Why this step? Multiplying by $4$ removes the fraction without changing the solution.

Answer. $x=6$ .

Check. $\dfrac{3(6)+2}{4}=5$ , and $6-1=5$ .

Worked example 2 changes only the coefficient on the right from $1$ to $2$ . The solution is now a fraction, which is a perfectly good exact answer.

Worked example 2: a fractional solution

Question. Solve $\dfrac{3x+2}{4}=2x-1$ .

Plan. Clear the denominator and collect terms before dividing.

\begin{aligned} 3x+2&=8x-4 && \text{multiply both sides by }4 \\ 6&=5x && \text{collect terms} \\ x&=\frac65 && \text{divide by }5 \end{aligned}

Why this step? The whole expression $2x-1$ is multiplied by $4$ , giving $8x-4$ .

Answer. $x=\frac65$ .

Check. Both sides equal $\frac75$ when $x=\frac65$ .

Equations also compare real quantities. If two hire charges are £ $4+3t$ and £ $1+4t$ for $t$ hours, equal charges mean $4+3t=1+4t$ , so $t=3$ . Both charges are then £ $13$ . Writing the comparison correctly comes before solving it.

Watch out

Wrong: multiplying $\frac{3x+2}{4}=2x-1$ by $4$ gives $3x+2=8x-1$ . Right: it gives $3x+2=8x-4$ ; the multiplier acts on the entire side.

The same balancing operations work with a different numerator. They are not supplied as a solving procedure on the formulae sheet: learn them.

Solving by factorisation

Factorisation changes a quadratic equation into a product equal to zero, which gives a small choice of linear equations.

Quadratic equation

A quadratic equation simplifies to $ax^2+bx+c=0$ , where $a\ne0$ . For example, $2x^2-x-6=0$ is quadratic. An expression such as $2x^2-x-6$ is not an equation until an equality is stated.

You already know how to factorise. The new step is using what a zero product tells you: if neither factor were zero, their product could not be zero.

The zero-product rule

For real numbers $A$ and $B$ ,

AB=0\quad\Longleftrightarrow\quad A=0\text{ or }B=0

The product must equal zero before the rule can be used.
Not on the formulae sheet: learn it.

Rearrange first, factorise second and solve each factor last. This keeps an expression, its factors and the equation's solutions distinct.

Worked example 3: two linear factors

Question. Solve $x^2-x=6$ by factorising.

Plan. Make one side zero, factorise and set each factor equal to zero.

\begin{aligned} x^2-x-6&=0 && \text{subtract }6 \\ (x-3)(x+2)&=0 && \text{factorise} \\ x-3=0&\quad\text{or}\quad x+2=0 && \text{zero-product rule} \end{aligned}

Why this step? A product equal to zero must have a zero factor, so both factors give possible solutions.

Answer. $x=3$ or $x=-2$ .

Check. $3^2-3=6$ and $(-2)^2-(-2)=6$ .

Worked example 4 changes only the coefficient of $x^2$ from $1$ to $2$ . One factor now gives a fractional root.

Worked example 4: a non-monic quadratic

Question. Solve $2x^2-x=6$ by factorising.

Plan. Rearrange to zero, then factorise the non-monic quadratic.

\begin{aligned} 2x^2-x-6&=0 && \text{subtract }6 \\ (2x+3)(x-2)&=0 && \text{factorise} \\ 2x+3=0&\quad\text{or}\quad x-2=0 && \text{zero-product rule} \end{aligned}

Why this step? The factor $2x+3$ is zero at $-\frac32$ , not at $-3$ .

Answer. $x=-\frac32$ or $x=2$ .

Check. $2\left(-\frac32\right)^2-\left(-\frac32\right)=6$ , and $2(2)^2-2=6$ .

Now return briefly to the rectangle. If its width is $w$ centimetres, its length is $w+3$ centimetres, so

w(w+3)=40\quad\Longrightarrow\quad (w-5)(w+8)=0

The algebra gives $w=5$ or $w=-8$ . A width must be positive, so only $5\,\mathrm{cm}$ is possible; the length is $8\,\mathrm{cm}$ . Both roots solve the equation, but only one fits the context. That is the answer to the opening puzzle.

Watch out

Wrong: divide $x(x-4)=0$ by $x$ and keep only $x=4$ . Dividing by $x$ assumes $x\ne0$ and loses the solution $x=0$ . Right: use the zero-product rule to keep both $0$ and $4$ .

The next equation changes the middle coefficient. Write the solutions as values of $x$ , rather than leaving two factors on the answer line.

Solving by completing the square

Completing the square is especially useful when a quadratic does not factorise neatly: isolate a squared bracket, then take both square roots.

You already know

Completing the square rewrites a quadratic without changing its value. For example, $x^2+6x+2=(x+3)^2-7$ . If the leading coefficient is not $1$ , divide the equation by that coefficient first.

There is a useful surprise hidden in this form. The two solutions of $(x+3)^2=7$ lie equally far either side of $-3$ , even though both are irrational. A completed square displays their midpoint before you calculate either solution.

Taking square roots

For real $h$ and $k>0$ ,

(x-h)^2=k\quad\Longrightarrow\quad x=h\pm\sqrt{k}

If $k=0$ , there is one repeated solution $x=h$ . If $k<0$ , there are no real solutions, because a real square cannot be negative.
Not on the formulae sheet: learn it.

The sign belongs to the square root after the squared bracket has been isolated. It does not mean changing every sign in the answer.

Worked example 5: irrational roots

Question. Solve $x^2+6x+2=0$ by completing the square. Give exact answers.

Plan. Rewrite the quadratic as a square minus a constant, then take both roots.

\begin{aligned} (x+3)^2-7&=0 && \text{complete the square} \\ (x+3)^2&=7 && \text{add }7 \\ x+3&=\pm\sqrt7 && \text{both square roots} \\ x&=-3\pm\sqrt7 && \text{subtract }3 \end{aligned}

Why this step? Both $\sqrt7$ and $-\sqrt7$ square to $7$ , so omitting either sign loses a solution.

Answer. $x=-3+\sqrt7$ or $x=-3-\sqrt7$ .

Check. For either value, $(x+3)^2-7=(\pm\sqrt7)^2-7=0$ .

Worked example 6 changes only the leading coefficient from $1$ to $2$ . Dividing first makes the same completing-square method available.

Worked example 6: divide before completing

Question. Solve $2x^2+6x+2=0$ by completing the square. Give exact answers.

Plan. Divide by $2$ , complete the square and simplify the square root.

\begin{aligned} x^2+3x+1&=0 && \text{divide by }2 \\ \left(x+\frac32\right)^2-\frac54&=0 && \text{complete the square} \\ x+\frac32&=\pm\frac{\sqrt5}{2} && \text{take both roots} \\ x&=\frac{-3\pm\sqrt5}{2} && \text{subtract }\frac32 \end{aligned}

Why this step? Half of the coefficient of $x$ is $\frac32$ ; its square is $\frac94$ , so the adjustment is $1-\frac94=-\frac54$ .

Answer. $x=\frac{-3+\sqrt5}{2}$ or $x=\frac{-3-\sqrt5}{2}$ .

Check. Each root makes $x+\frac32=\pm\frac{\sqrt5}{2}$ , whose square is $\frac54$ .

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2.14 - Solving linear and quadratic equations

Keeping an equation balanced

Solving by factorisation

Solving by completing the square

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