1.1 - Solving quadratic equations | Cambridge A-Level Mathematics

A quadratic equation may have two different real solutions, a repeated solution, or no real solution. You will solve it by factorising, completing the square and using the quadratic formula, then choose a method that fits the equation. Exact answers will let you check the algebra without rounding.

Zero products

Why can changing the constant in $x^2-6x+8=0$ by just $1$ change two answers into one? The key is to rewrite the left side as a product.

For ordinary real numbers, a product can be zero only if at least one of its factors is zero. Thus $(x-2)(x-4)=0$ gives $x-2=0$ or $x-4=0$ . It does not require both factors to be zero at the same input.

Root and repeated root

A root of an equation is a value that satisfies it. A repeated root occurs when the same linear factor appears twice in a quadratic: $(x-r)^2=0$ has the single distinct solution $x=r$ , counted twice as a root.

Bring every term to one side before using the zero-product property. For example, $x^2=5x$ becomes $x(x-5)=0$ , giving $x=0$ or $x=5$ . Dividing by $x$ would lose the zero solution.

Worked example 1: two different roots

Question. Solve $x^2-6x+8=0$ .

Plan. Find two numbers whose product is $8$ and whose sum is $-6$ , then set each factor equal to zero.

\begin{aligned} x^2-6x+8&=(x-2)(x-4) &&\text{expand to verify}\\ (x-2)(x-4)&=0\\ x&=2\text{ or }4 &&\text{one factor must vanish} \end{aligned}

Answer. $x=2$ or $x=4$ .

Check. The original expression is $4-12+8=0$ at $x=2$ and $16-24+8=0$ at $x=4$ .

Change only the constant from $8$ to $9$ . The two factors now coincide.

Worked example 2: a repeated root

Question. Solve $x^2-6x+9=0$ .

Plan. Recognise a perfect square, then find when its base is zero.

\begin{aligned} x^2-6x+9&=(x-3)^2\\ (x-3)^2&=0\\ x-3&=0 &&\text{only zero has square zero} \end{aligned}

Answer. $x=3$ , a repeated root.

Check. $3^2-6(3)+9=0$ . Both factors $x-3$ vanish at this same value.

For a non-monic quadratic, check the coefficient of $x^2$ as well as the constant: $2x^2+x-3=(2x+3)(x-1)$ . Expansion checks all three coefficients.

Watch out

From $(x-2)(x-4)=0$ , write $x=2$ or $x=4$ . From $(x-2)(x-4)=6$ , you cannot set either factor to $6$ : first rearrange the equation to have zero on one side.

Try a different repeated factor.

Completing a square to solve

Completing the square is useful even when integer factorisation is unavailable. Once you reach $(x-h)^2=d$ , the sign of $d$ tells you what to do.

If $d>0$ , the base may be positive or negative: $x-h=\pm\sqrt d$ . If $d=0$ , there is one repeated root. If $d<0$ , there is no real solution because a real square cannot be negative.

Solving an isolated square

For real $h$ and $d\ge0$ ,

(x-h)^2=d\quad\Longrightarrow\quad x=h\pm\sqrt d.

When $d=0$ , the two signs give the same value. Not in the formula booklet: learn it.

Keep surds exact. Recall that $\sqrt{20}=2\sqrt5$ , while $\sqrt{a+b}$ is generally not $\sqrt a+\sqrt b$ . If an exact answer has a simple surd denominator, rationalise it: $1/\sqrt2=\sqrt2/2$ , obtained by multiplying numerator and denominator by $\sqrt2$ .

Worked example 3: two exact surd roots

Question. Solve $x^2+4x-1=0$ , giving exact answers.

Plan. Complete the square and take both square roots.

\begin{aligned} x^2+4x-1&=(x+2)^2-5\\ (x+2)^2&=5 &&\text{isolate the square}\\ x+2&=\pm\sqrt5\\ x&=-2\pm\sqrt5 \end{aligned}

Answer. $x=-2+\sqrt5$ or $x=-2-\sqrt5$ .

Check. Put $x=-2+s$ , where $s=\pm\sqrt5$ . Direct expansion of the original gives $4-4s+s^2-8+4s-1=s^2-5=0$ for both signs.

Now change only the constant from $-1$ to $5$ . Watch what happens to the isolated square.

Worked example 4: no real solution

Question. Solve $x^2+4x+5=0$ for real $x$ .

Plan. Complete the same square and examine the required value of that square.

\begin{aligned} x^2+4x+5&=(x+2)^2+1\\ (x+2)^2&=-1 \end{aligned}

A real square is at least zero, so the last equation is impossible.

Answer. There is no real solution.

Check. The original expression has minimum value $1$ at $x=-2$ , so it never reaches zero.

The symbol $\sqrt5$ denotes the non-negative square root. The $\pm$ enters because we are solving an equation, not because the square-root symbol has two values.

Watch out

$(x+2)^2=5$ does not give only $x+2=\sqrt5$ . Include $x+2=-\sqrt5$ as well; otherwise one valid root disappears.

Use exact surds in this check.

The quadratic formula

The quadratic formula packages the completing-square method into one expression. It is particularly useful when the coefficient of $x^2$ is not $1$ and factors are not easy to spot.

Quadratic formula

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

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

In the formula booklet.

First put the equation in this zero-right-side form, then identify the signed coefficients. The whole numerator is divided by $2a$ . A negative value under the square root gives no real answer; the next lesson examines that expression in its own right.

Worked example 5: a non-monic quadratic

Question. Solve $2x^2+3x-1=0$ , giving exact answers.

Plan. Use $a=2$ , $b=3$ , $c=-1$ in the supplied formula.

\begin{aligned} x&=\frac{-3\pm\sqrt{3^2-4(2)(-1)}}{2(2)}\\ &=\frac{-3\pm\sqrt{17}}4 \end{aligned}

Answer. $x=(-3+\sqrt{17})/4$ or $x=(-3-\sqrt{17})/4$ .

Check. The original expression equals $[(4x+3)^2-17]/8$ . At either proposed root, $4x+3=\pm\sqrt{17}$ , so this expression is zero. The approximate roots $0.281$ and $-1.78$ also have opposite signs, consistent with the negative constant and positive leading coefficient.

Change only the sign of the linear coefficient. This changes $-b$ , while $b^2$ stays the same.

Worked example 6: a negative linear coefficient

Question. Solve $2x^2-3x-1=0$ , giving exact answers.

Plan. Keep the negative sign in $b=-3$ when substituting.

\begin{aligned} x&=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(-1)}}4\\ &=\frac{3\pm\sqrt{17}}4 \end{aligned}

Answer. $x=(3+\sqrt{17})/4$ or $x=(3-\sqrt{17})/4$ .

Check. Replacing $x$ by $-x$ changes the previous equation into this one. Negating its two roots gives exactly this pair. Directly, $2x^2-3x-1=[(4x-3)^2-17]/8=0$ at both answers.

Choose a method by looking at the structure before calculating.

What you can see	An efficient first choice
Simple linear factors	Factorise and use a zero product
A square is already present or easy to make	Isolate or complete the square
No convenient factors or square	Use the quadratic formula

The opening equations have simple factors: one has roots $2,4$ , and adding $1$ makes the repeated root $3$ . The methods agree; their usefulness depends on what the expression reveals.

Exam tip

Show the substituted formula or the algebra that produces the roots. An unsupported calculator answer does not show the necessary working. If exact answers are requested, keep the surd rather than replacing it by a decimal.

Make the coefficient choices visible in your solution.

Quick Recap

Make one side zero before factorising.
An isolated square gives both signs when its value is positive, one root when zero and no real root when negative.
Keep the signs of $a$ , $b$ and $c$ when using the quadratic formula.
Substitute answers back or use an equivalent identity to check them.

Changing the constant from $8$ to $9$ joins the two factors into the same factor: two distinct roots become one repeated root.

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1.1 - Solving quadratic equations

Zero products

Completing a square to solve

The quadratic formula

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