E2.5 - Solving quadratic equations | Cambridge IGCSE Mathematics

Quadratic equations can have two, one or no real solutions. In this Extended lesson, you will solve by factorisation, completing the square and the supplied quadratic formula, keep exact roots when needed, and decide which solutions fit a situation.

Solving by factorisation

Can changing just one number turn an equation with two real solutions into one with none? We will use the equation $x^2+6x+c=0$ to answer that question after building three solving methods.

A quadratic equation can be written $ax^2+bx+c=0$ , where $a\ne0$ . Its solutions are also called roots. Factorisation is often the shortest method when the quadratic splits into simple factors.

Zero-product rule

For real numbers $u$ and $v$ , if $uv=0$ , then $u=0$ or $v=0$ , or both. If neither factor were zero, their product could not be zero. For example, $(x-2)(x+3)=0$ requires $x-2=0$ or $x+3=0$ .

The right side must be zero before you use this rule. A product equal to six does not mean one factor is six or zero.

Worked example 1: a quadratic with leading coefficient one

Question. Solve $x^2+x-6=0$ .

Plan. Factorise, then set each factor equal to zero.

\begin{aligned} x^2+x-6&=(x+3)(x-2) && \text{product }-6,\ \text{sum }1\\ (x+3)(x-2)&=0 && \text{use the equation}\\ x+3=0&\quad\text{or}\quad x-2=0 && \text{zero-product rule}\\ x=-3&\quad\text{or}\quad x=2 && \text{solve each linear equation} \end{aligned}

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

Check. $(-3)^2+(-3)-6=0$ , and $2^2+2-6=0$ .

Change only the leading coefficient from one to two. The factor involving $2x$ now produces a fractional root.

Worked example 2: a fractional root from a factor

Question. Solve $2x^2+x-6=0$ .

Plan. Factorise with a $2x$ term, then solve both factor equations.

\begin{aligned} 2x^2+x-6&=(2x-3)(x+2) && \text{middle terms }4x-3x\\ (2x-3)(x+2)&=0 && \text{replace the quadratic}\\ 2x-3=0&\quad\text{or}\quad x+2=0 && \text{one factor must be zero}\\ x=\frac32&\quad\text{or}\quad x=-2 && \text{solve both} \end{aligned}

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

Check. $2(3/2)^2+3/2-6=9/2+3/2-6=0$ , and $2(-2)^2-2-6=0$ .

If you start with $x^2=5x$ , first write $x^2-5x=0$ , then $x(x-5)=0$ . The roots are zero and five. Dividing by $x$ would wrongly discard the zero root, because that division assumes $x\ne0$ .

Watch out

From $(2x-3)(x+2)=0$ , the answers are not $3$ and $-2$ . Solve $2x-3=0$ to get $x=3/2$ . The coefficient inside a factor still matters.

A fractional root is as valid as an integer root; the original equation decides whether it works.

Solving by completing the square

Completing the square rewrites a quadratic as a square plus or minus a number. It exposes how many real solutions are possible, including when simple integer factorisation does not work.

Recall $(x+h)^2=x^2+2hx+h^2$ . For $x^2+6x$ , half the coefficient of $x$ is three, so add and subtract nine:

x^2+6x+c=(x+3)^2+c-9.

This is an identity, valid for every real $x$ . To solve an equation, isolate the square and then take both square-root branches.

Worked example 3: two integer roots from a square

Question. Solve $x^2+6x+5=0$ by completing the square.

Plan. Make $(x+3)^2$ , isolate it, then solve the two branches.

\begin{aligned} (x+3)^2-9+5&=0 && \text{complete the square}\\ (x+3)^2&=4 && \text{add four}\\ x+3&=\pm2 && \text{take both roots}\\ x&=-3\pm2 && \text{subtract three} \end{aligned}

Answer. $x=-1$ or $x=-5$ .

Check. $1-6+5=0$ and $25-30+5=0$ in the original equation.

Change only the constant from five to two. The square is now seven, so the exact roots contain a surd.

Worked example 4: keeping exact surd roots

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

Plan. Use the same completed square, then retain $\sqrt7$ rather than rounding it.

\begin{aligned} (x+3)^2-9+2&=0 && \text{complete the square}\\ (x+3)^2&=7 && \text{add seven}\\ x+3&=\pm\sqrt7 && \text{two real branches}\\ x&=-3\pm\sqrt7 && \text{subtract three} \end{aligned}

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

Check. For either sign, $(x+3)^2=(\pm\sqrt7)^2=7$ . Hence the original expression, $(x+3)^2-7$ , equals zero.

Now the opening question can be answered precisely:

x^2+6x+c=0\quad\Longleftrightarrow\quad (x+3)^2=9-c.

For $c<9$ , the right side is positive and there are two distinct real roots. For $c=9$ , it is zero and the sole distinct root is $x=-3$ . The factor $(x+3)$ occurs twice, so this is a repeated root. For $c>9$ , there is no real root, because a real square cannot be negative. For example, $c=10$ would require $(x+3)^2=-1$ .

If the leading coefficient is not one, first divide the equation by its non-zero leading coefficient. You can then complete the square using half the new coefficient of $x$ .

Watch out

From $(x+3)^2=7$ , writing only $x=-3+\sqrt7$ loses a solution. From $(x+3)^2=-1$ , writing $x=-3\pm1$ invents solutions. Decide whether the isolated square is positive, zero or negative.

An estimate can check a surd root without replacing its exact form: $2<\sqrt7<3$ , so $-3+\sqrt7$ lies between $-1$ and zero.

Using the quadratic formula

The quadratic formula packages completing the square into a method for any quadratic. First write the equation with zero on one side. Read off $a$ , $b$ and $c$ , including their signs, from $ax^2+bx+c=0$ .

Quadratic formula

For $ax^2+bx+c=0$ , where $a\ne0$ ,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

For real solutions, $b^2-4ac\ge0$ . A positive value gives two distinct real roots; zero gives a repeated real root; a negative value gives no real roots.

Given in the Extended List of formulas.

Why does it work? Multiply $ax^2+bx+c=0$ by $4a$ , move $4ac$ to the right and add $b^2$ :

\begin{aligned} 4a^2x^2+4abx+b^2&=b^2-4ac && \text{complete a square}\\ (2ax+b)^2&=b^2-4ac && \text{factor the square}\\ 2ax+b&=\pm\sqrt{b^2-4ac} && \text{when the right side is non-negative} \end{aligned}

Subtract $b$ and divide by $2a$ to obtain the formula. If $a=0$ , this division is impossible and the equation is not quadratic.

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E2.5 - Solving quadratic equations

Solving by factorisation

Solving by completing the square

Using the quadratic formula

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