Mathematics 9709 · Pure Mathematics 1

The quadratic formula

Sometimes the roots from the calculator do not make a clean factorised line. Or the question wants an exact surd answer. In both cases, use the formula.
It always works, even when the numbers are awkward.

The formula

For $ax^2+bx+c=0$ , the roots are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ . This is completing the square, done once for every quadratic.

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

The $\pm$ means do it twice. Once you add $\sqrt{b^2-4ac}$ , and once you subtract it. This is why a quadratic usually has two answers.

Why this is just completing the square

This formula is completing the square from the last topic. It is done once, in general, on $ax^2+bx+c=0$ . Because the work is done for you, you never have to repeat it. You just read off three numbers and the answer comes out. The square root and the $\pm$ are the “take the square root of both sides” step. That step is where the two answers come from.

When to switch to the formula

Do not guess at the start. Let the roots tell you. Get to $=0$ , read both roots, and try to write them as a clean $(x-r_1)(x-r_2)$ bracket line.

If the roots do not make a clean bracket line, switch to the formula.

Two that look the same:

$x^2 - 5x + 6 = 0 \qquad\text{vs}\qquad x^2 - 4x + 1 = 0$

Left: the roots are whole numbers, $2,\,3$ . So the bracket is easy to write, $(x-2)(x-3)=0$ . That line is your method.
Right: the roots are messy ( $\approx 3.73,\ 0.27$ ). There is nothing clean to write, so use the formula.

Worked example

So use the formula on $x^2-4x+1=0$

Read off with signs: $a=1,\ b=-4,\ c=1$ .
Discriminant first (this is where most sign mistakes happen): $b^2-4ac=(-4)^2-4(1)(1)=16-4=12$ .
Substitute ( $-b=+4$ , $2a=2$ ) — this line is your method mark, exactly like the bracket line was: $x=\dfrac{4\pm\sqrt{12}}{2}$ .
Tidy the surd ( $\sqrt{12}=2\sqrt3$ ) and cancel the shared 2: $x=\dfrac{4\pm2\sqrt3}{2}=2\pm\sqrt3$ .

Answer

x=2\pm\sqrt3

Those are the exact surds. They are the same numbers the calculator rounded to $3.73$ and $0.27$ . The formula gave you two things: a valid shown method, and the exact answer that a decimal cannot give.

Reading off a, b and c

Line up against $ax^2+bx+c$ and keep each sign attached to its number.

$a$ sits in front of $x^2$ .
$b$ sits in front of $x$ .
$c$ is the number on its own.
A minus belongs to the number after it — so in $3x^2-5x+1=0$ , $a=3,\ b=-5,\ c=1$ .

\underbrace{6}_{a}\,x^2\;\underbrace{-\,3}_{b}\,x\;\underbrace{-\,2}_{c}=0

Match each coefficient to its slot — and keep the sign attached.

Why the sign matters so much

In $2x^2-3x-2$ the $b$ is $-3$ (with the minus), and $c$ is $-2$ . The formula squares $b$ and uses $-b$ . So if you drop one minus here, a later sign goes wrong and the whole answer is wrong. This is the most common mistake in the topic.

Working through the formula

The steps are always the same: read off → work out $b^2-4ac$ → substitute → tidy up.

Read off $a,\ b,\ c$ with signs.
Work out the inside of the root, $b^2-4ac$ — the part most likely to go wrong.
Substitute into $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ .
Split the $\pm$ and tidy (simplify the surd, or round).

Check: it gives the same answer as factorising

Try it on one that does factorise. For $2x^2+7x+3=0$ , $a=2,\ b=7,\ c=3$ and $b^2-4ac=(7)^2-4(2)(3)=49-24=25$ . This is a perfect square ( $\sqrt{25}=5$ ), and that is why it factorises. So $x=\dfrac{-7\pm 5}{4}$ gives $x=-\tfrac12$ or $x=-3$ . This is the same as $(2x+1)(x+3)=0$ , as it should be.

Worked example

A standard case — solve $2x^2-4x-3=0$

Read off: $a=2,\ b=-4,\ c=-3$ .
Root first: $b^2-4ac=(-4)^2-4(2)(-3)=16+24=40$ .
Substitute ( $-b=+4$ ): $x=\dfrac{4\pm\sqrt{40}}{4}$ .
Tidy the surd, then cancel the shared 2: $x=\dfrac{4\pm 2\sqrt{10}}{4}=\dfrac{2\pm\sqrt{10}}{2}$ .

Answer

x=\dfrac{2+\sqrt{10}}{2}\quad\text{or}\quad x=\dfrac{2-\sqrt{10}}{2}

Show the full working

The $-4(2)(-3)$ is a minus times a minus, so it becomes a plus. That gives the $+24$ : $b^2-4ac=16-(-24)=16+24=40$ .

Substituting with $-b=-(-4)=+4$ and $2a=4$ : $x=\dfrac{-(-4)\pm\sqrt{40}}{2(2)}=\dfrac{4\pm\sqrt{40}}{4}$ .

Since $40=4\times 10$ and $\sqrt{4}=2$ , $\sqrt{40}=2\sqrt{10}$ , giving $x=\dfrac{4\pm 2\sqrt{10}}{4}$ ; top and bottom share a factor of 2, so $x=\dfrac{2\pm\sqrt{10}}{2}$ . As decimals $x\approx 2.58$ or $x\approx -0.581$ (3 s.f.).

Your turn— tap to reveal the worked answer (quick check)

Find only the value inside the root for $x^2-2x-5=0$ — just $b^2-4ac$ .

$a=1,\ b=-2,\ c=-5$ , so $b^2-4ac=(-2)^2-4(1)(-5)=4+20$ .

Answer

b^2-4ac=24

Worked example

Exam level — solve $6x^2-3x-2=0$ to 3 significant figures

Read off: $a=6,\ b=-3,\ c=-2$ .
Root first: $b^2-4ac=(-3)^2-4(6)(-2)=9+48=57$ .
Substitute ( $-b=+3$ ): $x=\dfrac{3\pm\sqrt{57}}{12}$ .
Split and round: $\sqrt{57}\approx 7.5498$ .

Answer

x=0.879\quad\text{or}\quad x=-0.379

Show the two cases

$57$ is not a perfect square, so the answer will not be tidy. That is why the question asks for 3 s.f. From $x=\dfrac{3\pm\sqrt{57}}{2(6)}=\dfrac{3\pm\sqrt{57}}{12}$ , split the $\pm$ :

$x=\dfrac{3+7.5498}{12}=\dfrac{10.5498}{12}\approx 0.879$

$x=\dfrac{3-7.5498}{12}=\dfrac{-4.5498}{12}\approx -0.379$

Exact (surd) answers

If it says “exact” or asks for the form $p\pm q\sqrt{n}$ , keep the surd in. Do not use a calculator.

Get to $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ as usual.
Simplify the surd with $\sqrt{a\,b}=\sqrt{a}\,\sqrt{b}$ , pulling out any perfect square inside.
Cancel any factor shared by every term — then stop.

Worked example: x² + 4x − 1 = 0

$a=1,\ b=4,\ c=-1$ , so $b^2-4ac=(4)^2-4(1)(-1)=16+4=20$ and $x=\dfrac{-4\pm\sqrt{20}}{2}$ . Since $20=4\times 5$ , $\sqrt{20}=2\sqrt5$ , giving $x=\dfrac{-4\pm 2\sqrt5}{2}$ ; cancel the shared 2 from every term to get the exact answer $x=-2\pm\sqrt5$ .

Where the marks go

Divide everything by $2a$ . Check the sign of $b$ . And cancel from every term.

Common mistake

The whole top line

-b\pm\sqrt{\,\cdots}

is divided by

2a

, not just the square root. Draw the fraction bar long enough to sit under everything. And check the sign of

b

: with

b=-3

-b=+3

. This is the most common one-mark mistake in this topic.

Common mistake

When you cancel a common factor, cancel it from every term:

\frac{-4\pm 2\sqrt5}{2}=-2\pm\sqrt5

, not

-2\pm 2\sqrt5

. And “exact” means keep the surd in. Only round when the question asks for significant figures or decimal places.

Now you try

Solve $x^2+6x+4=0$ , giving exact answers. (9709-style)

Your turn— tap to reveal the worked answer (leave it in surd form)

Read off $a=1,\ b=6,\ c=4$ .
Root: $b^2-4ac=(6)^2-4(1)(4)=36-16=20$ .
Substitute: $x=\dfrac{-6\pm\sqrt{20}}{2}$ .
Tidy $\sqrt{20}=2\sqrt5$ and cancel the 2: $x=\dfrac{-6\pm 2\sqrt5}{2}=-3\pm\sqrt5$ .

Answer

x=-3\pm\sqrt5

PreviousCompleting the square NextThe quadratic graph

The quadratic formula

Sometimes the roots from the calculator do not make a clean factorised line. Or the question wants an exact surd answer. In both cases, use the formula.
It always works, even when the numbers are awkward.

The formula

For $ax^2+bx+c=0$ , the roots are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ . This is completing the square, done once for every quadratic.

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

The $\pm$ means do it twice. Once you add $\sqrt{b^2-4ac}$ , and once you subtract it. This is why a quadratic usually has two answers.

Why this is just completing the square

When to switch to the formula

Do not guess at the start. Let the roots tell you. Get to $=0$ , read both roots, and try to write them as a clean $(x-r_1)(x-r_2)$ bracket line.

If the roots do not make a clean bracket line, switch to the formula.

Two that look the same:

$x^2 - 5x + 6 = 0 \qquad\text{vs}\qquad x^2 - 4x + 1 = 0$

Left: the roots are whole numbers, $2,\,3$ . So the bracket is easy to write, $(x-2)(x-3)=0$ . That line is your method.
Right: the roots are messy ( $\approx 3.73,\ 0.27$ ). There is nothing clean to write, so use the formula.

Worked example

So use the formula on $x^2-4x+1=0$

Read off with signs: $a=1,\ b=-4,\ c=1$ .
Discriminant first (this is where most sign mistakes happen): $b^2-4ac=(-4)^2-4(1)(1)=16-4=12$ .
Substitute ( $-b=+4$ , $2a=2$ ) — this line is your method mark, exactly like the bracket line was: $x=\dfrac{4\pm\sqrt{12}}{2}$ .
Tidy the surd ( $\sqrt{12}=2\sqrt3$ ) and cancel the shared 2: $x=\dfrac{4\pm2\sqrt3}{2}=2\pm\sqrt3$ .

Answer

x=2\pm\sqrt3

Reading off a, b and c

Line up against $ax^2+bx+c$ and keep each sign attached to its number.

$a$ sits in front of $x^2$ .
$b$ sits in front of $x$ .
$c$ is the number on its own.
A minus belongs to the number after it — so in $3x^2-5x+1=0$ , $a=3,\ b=-5,\ c=1$ .

\underbrace{6}_{a}\,x^2\;\underbrace{-\,3}_{b}\,x\;\underbrace{-\,2}_{c}=0

Match each coefficient to its slot — and keep the sign attached.

Why the sign matters so much

Working through the formula

The steps are always the same: read off → work out $b^2-4ac$ → substitute → tidy up.

Read off $a,\ b,\ c$ with signs.
Work out the inside of the root, $b^2-4ac$ — the part most likely to go wrong.
Substitute into $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ .
Split the $\pm$ and tidy (simplify the surd, or round).

Check: it gives the same answer as factorising

Worked example

A standard case — solve $2x^2-4x-3=0$

Read off: $a=2,\ b=-4,\ c=-3$ .
Root first: $b^2-4ac=(-4)^2-4(2)(-3)=16+24=40$ .
Substitute ( $-b=+4$ ): $x=\dfrac{4\pm\sqrt{40}}{4}$ .
Tidy the surd, then cancel the shared 2: $x=\dfrac{4\pm 2\sqrt{10}}{4}=\dfrac{2\pm\sqrt{10}}{2}$ .

Answer

x=\dfrac{2+\sqrt{10}}{2}\quad\text{or}\quad x=\dfrac{2-\sqrt{10}}{2}

Show the full working

The $-4(2)(-3)$ is a minus times a minus, so it becomes a plus. That gives the $+24$ : $b^2-4ac=16-(-24)=16+24=40$ .

Substituting with $-b=-(-4)=+4$ and $2a=4$ : $x=\dfrac{-(-4)\pm\sqrt{40}}{2(2)}=\dfrac{4\pm\sqrt{40}}{4}$ .

Your turn— tap to reveal the worked answer (quick check)

Find only the value inside the root for $x^2-2x-5=0$ — just $b^2-4ac$ .

$a=1,\ b=-2,\ c=-5$ , so $b^2-4ac=(-2)^2-4(1)(-5)=4+20$ .

Answer

b^2-4ac=24

Worked example

Exam level — solve $6x^2-3x-2=0$ to 3 significant figures

Read off: $a=6,\ b=-3,\ c=-2$ .
Root first: $b^2-4ac=(-3)^2-4(6)(-2)=9+48=57$ .
Substitute ( $-b=+3$ ): $x=\dfrac{3\pm\sqrt{57}}{12}$ .
Split and round: $\sqrt{57}\approx 7.5498$ .

Answer

x=0.879\quad\text{or}\quad x=-0.379

Show the two cases

$57$ is not a perfect square, so the answer will not be tidy. That is why the question asks for 3 s.f. From $x=\dfrac{3\pm\sqrt{57}}{2(6)}=\dfrac{3\pm\sqrt{57}}{12}$ , split the $\pm$ :

$x=\dfrac{3+7.5498}{12}=\dfrac{10.5498}{12}\approx 0.879$

$x=\dfrac{3-7.5498}{12}=\dfrac{-4.5498}{12}\approx -0.379$

Exact (surd) answers

If it says “exact” or asks for the form $p\pm q\sqrt{n}$ , keep the surd in. Do not use a calculator.

Get to $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ as usual.
Simplify the surd with $\sqrt{a\,b}=\sqrt{a}\,\sqrt{b}$ , pulling out any perfect square inside.
Cancel any factor shared by every term — then stop.

Worked example: x² + 4x − 1 = 0

Where the marks go

Divide everything by $2a$ . Check the sign of $b$ . And cancel from every term.

Common mistake

The whole top line

-b\pm\sqrt{\,\cdots}

is divided by

2a

, not just the square root. Draw the fraction bar long enough to sit under everything. And check the sign of

b

: with

b=-3

-b=+3

. This is the most common one-mark mistake in this topic.

Common mistake

When you cancel a common factor, cancel it from every term:

\frac{-4\pm 2\sqrt5}{2}=-2\pm\sqrt5

, not

-2\pm 2\sqrt5

. And “exact” means keep the surd in. Only round when the question asks for significant figures or decimal places.

Now you try

Solve $x^2+6x+4=0$ , giving exact answers. (9709-style)

Your turn— tap to reveal the worked answer (leave it in surd form)

Read off $a=1,\ b=6,\ c=4$ .
Root: $b^2-4ac=(6)^2-4(1)(4)=36-16=20$ .
Substitute: $x=\dfrac{-6\pm\sqrt{20}}{2}$ .
Tidy $\sqrt{20}=2\sqrt5$ and cancel the 2: $x=\dfrac{-6\pm 2\sqrt5}{2}=-3\pm\sqrt5$ .

Answer

x=-3\pm\sqrt5