Quadratic Formula

What you'll learn

How to recognise a quadratic equation and write it in standard form.
How to substitute correctly into the quadratic formula.
How to give answers as decimals or exact surds.
How to use the formula in area and Pythagoras problems.

Before the formula: standard form

To use the quadratic formula, the equation must be written with everything on one side and zero on the other side.

Definition

Quadratic equation

A quadratic equation is an equation where the highest power of the variable is squared, such as $x^2$ . The standard form is $ax^2+bx+c=0$ , where $a$ , $b$ and $c$ are coefficients: the numbers multiplying the terms. You must have $\neq 0$ . A solution or root is a value of $x$ that makes the equation true.

A quadratic graph is called a parabola, which is a U-shaped curve. The roots are where the graph crosses the x-axis.

Parabola showing roots as x-intercepts

Key Idea

Why standard form matters

The formula only works when the equation is in the form $ax^2+bx+c=0$ . Rearrange first, then identify $a$ , $b$ and $c$ with their signs.

Example

Writing a quadratic in standard form

Solve later if needed, but first rewrite $4x^2=7x+2$ in standard form and identify $a$ , $b$ and $c$ .

Move every term to the left-hand side.
$4x^2-7x-2=0$
Compare with $ax^2+bx+c=0$ .
The coefficients are $a = 4$ , $b = - 7$ and $c = - 2$ .

Common Mistake

Losing the signs

If the equation is $4x^2-7x-2=0$ , then $b = - 7$ and $c = - 2$ . The minus signs belong to the coefficients.

The quadratic formula

The quadratic formula solves any quadratic equation in standard form, even when factorising is awkward or impossible.

Definition

Quadratic formula

For $ax^2+bx+c=0$ , the solutions are

x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

The symbol $±\pm$ means “plus or minus”, so it usually gives two answers.

The expression under the square root, $b^2-4ac$ , is called the discriminant. It helps decide what kind of answers you get.

If it is positive, there are two real roots.
If it is zero, there is one repeated root.
If it is negative, there are no real roots.

Common Mistake

When the formula does not apply

The quadratic formula is not for linear equations. If $a = 0$ , there is no $x^2$ term, so the equation is not quadratic.

Example

Solving a quadratic to 2 decimal places

Solve $3x^2+4x-5=0$ , giving your answers correct to 2 decimal places.

Identify the coefficients: $a = 3$ , $b = 4$ and $c = - 5$ .
Substitute into the formula.
$x=−4±42−4(3)(−5)2(3)x=\frac{-4\pm\sqrt{4^2-4(3)(-5)}}{2(3)}$
Simplify the discriminant.
$4^2-4(3)(-5)=16+60=76$
Write the exact formula result.
$x=−4±766x=\frac{-4\pm\sqrt{76}}{6}$
Work out both values.
$x=−4+766≈0.786x=−4−766≈−2.120\begin{aligned} x&=\frac{-4+\sqrt{76}}{6}\approx 0.786\\ x&=\frac{-4-\sqrt{76}}{6}\approx -2.120 \end{aligned}$
Round to 2 decimal places: $x = 0.79$ or $x = - 2.12$ .

Tip

Calculator brackets

Type the numerator in brackets, especially for the negative answer: (-4 - sqrt(76)) ÷ 6. This avoids accidentally dividing only part of the expression.

Rounding your answers

A question may ask for answers correct to a certain number of decimal places or significant figures.

Decimal places count digits after the decimal point.
Significant figures count from the first non-zero digit.

For example, 1.6489 to 3 significant figures is 1.65, while -0.8489 to 3 significant figures is -0.849.

Example

Rearranging first, then rounding to 3 significant figures

Solve $5x^2=4x+7$ , giving your answers correct to 3 significant figures.

Rearrange into standard form.
$5x^2-4x-7=0$
Identify the coefficients: $a = 5$ , $b = - 4$ and $c = - 7$ .
Substitute into the quadratic formula.
$x=−(−4)±(−4)2−4(5)(−7)2(5)x=\frac{-(-4)\pm\sqrt{(-4)^2-4(5)(-7)}}{2(5)}$
Simplify carefully.
$x=4±16+14010x=\frac{4\pm\sqrt{16+140}}{10}$
Work out both values.
$x=4+15610≈1.649x=4−15610≈−0.849\begin{aligned} x&=\frac{4+\sqrt{156}}{10}\approx 1.649\\ x&=\frac{4-\sqrt{156}}{10}\approx -0.849 \end{aligned}$
To 3 significant figures, $x = 1.65$ or $x = - 0.849$ .

Common Mistake

Rounding too early

Do not round the square root first unless the question asks you to. Keep the full calculator value until the final answer.

Exact answers using surds

Sometimes the question does not want a decimal. It may ask for an exact answer in a form like $a±bca\pm b\sqrt{c}$ .

Definition

Surd

A surd is an irrational root left in exact form, such as $2\sqrt{2}$ or $7\sqrt{7}$ . You should simplify it if it has a square factor, for example $72=62\sqrt{72}=6\sqrt{2}$ .

Example

Giving answers in exact surd form

Solve $x^2+10x+7=0$ , giving your answers in the form $a±bca\pm b\sqrt{c}$ .

Identify $a = 1$ , $b = 10$ and $c = 7$ .
Substitute into the formula.
$x=−10±102−4(1)(7)2(1)x=\frac{-10\pm\sqrt{10^2-4(1)(7)}}{2(1)}$
Simplify the discriminant.
$10^2-4(1)(7)=100-28=72$
Simplify the surd.
$72=36⋅2=62\sqrt{72}=\sqrt{36\cdot 2}=6\sqrt{2}$
Substitute this back into the answer.
$x=−10±622x=\frac{-10\pm 6\sqrt{2}}{2}$
Divide both terms in the numerator by 2.
$x=−5±32x=-5\pm 3\sqrt{2}$

Using the formula in geometry problems

In problem-solving questions, the hard part is often forming the quadratic equation. Once you have it, the formula is the same as before.

Key Idea

Set up first, solve second

For area or triangle problems, write an equation using the geometry first. Then expand, simplify to standard form, and solve.

The diagram below shows a typical compound shape made from two rectangles. To find the total area, add the area of each rectangle.

Compound rectilinear shape made from two rectangles with algebraic side lengths

Example

Area of a compound shape

A compound shape is made from two rectangles. The left rectangle has width $x - 2$ and height $2 x + 3$ . The right rectangle has width $2 x - 4$ and height 5. The total area is 46 cm². Find $x$ to 3 significant figures.

Write the area of each rectangle.
$(x - 2) (2 x + 3) + 5 (2 x - 4) = 46$
Expand each part.
$2x^2-x-6+10x-20=46$
Simplify and rearrange to standard form.
$2x^2+9x-72=0$
Use the quadratic formula with $a = 2$ , $b = 9$ and $c = - 72$ .
$x=−9±92−4(2)(−72)2(2)x=\frac{-9\pm\sqrt{9^2-4(2)(-72)}}{2(2)}$
Simplify.
$x=−9±6574x=\frac{-9\pm\sqrt{657}}{4}$
Work out both possible values: $x≈4.158x\approx 4.158$ or $x≈−8.658x\approx -8.658$ .
Reject the negative value because lengths such as $x - 2$ and $2 x - 4$ must be positive. So $x = 4.16$ to 3 significant figures.

Tip

Check physical meaning

In geometry, a negative algebraic solution may be mathematically correct but impossible for a length. Always check whether your answer makes the side lengths positive.

Right-angled triangle problems

For a right-angled triangle, the hypotenuse is the longest side, opposite the right angle. Pythagoras’ theorem says that

a^2+b^2=c^2

where $c$ is the hypotenuse.

Example

Using Pythagoras to form a quadratic

A right-angled triangle has shorter sides $x + 3$ and $x + 4$ , and hypotenuse $x + 6$ . Find the exact value of $x$ .

Use Pythagoras’ theorem.
$x+3)^2+(x+4)^2=(x+6)^2$
Expand both sides.
$x^2+6x+9+x^2+8x+16=x^2+12x+36$
Simplify and rearrange.
$x^2+2x-11=0$
Apply the quadratic formula.
$x=−2±22−4(1)(−11)2x=\frac{-2\pm\sqrt{2^2-4(1)(-11)}}{2}$
Simplify the surd.
$x=−2±482=−1±23x=\frac{-2\pm\sqrt{48}}{2}=-1\pm 2\sqrt{3}$
Only $x=−1+23x=-1+2\sqrt{3}$ is valid, because the other value would make some side lengths negative.

Exam technique

In the exam

Put the equation into $ax^2+bx+c=0$ before using the formula.
Write down $a$ , $b$ and $c$ , including negative signs.
Use brackets in your calculator, keep full accuracy until the end, and answer in the requested form: decimal, significant figures, or surd.
In geometry questions, reject any value that makes a length negative.

Self review

Check yourself

Can you rearrange an equation like $6x^2=5x+8$ into standard form and identify $a$ , $b$ and $c$ ?
When $b$ is negative, can you correctly work out $- b$ in the formula?
If an exact answer is requested, can you simplify a surd such as $80\sqrt{80}$ ?

Quadratic Formula

What you'll learn

How to recognise a quadratic equation and write it in standard form.
How to substitute correctly into the quadratic formula.
How to give answers as decimals or exact surds.
How to use the formula in area and Pythagoras problems.

Before the formula: standard form

To use the quadratic formula, the equation must be written with everything on one side and zero on the other side.

Definition

Quadratic equation

A quadratic graph is called a parabola, which is a U-shaped curve. The roots are where the graph crosses the x-axis.

Parabola showing roots as x-intercepts

Key Idea

Why standard form matters

The formula only works when the equation is in the form $ax^2+bx+c=0$ . Rearrange first, then identify $a$ , $b$ and $c$ with their signs.

Example

Writing a quadratic in standard form

Solve later if needed, but first rewrite $4x^2=7x+2$ in standard form and identify $a$ , $b$ and $c$ .

Move every term to the left-hand side.
$4x^2-7x-2=0$
Compare with $ax^2+bx+c=0$ .
The coefficients are $a = 4$ , $b = - 7$ and $c = - 2$ .

Common Mistake

Losing the signs

If the equation is $4x^2-7x-2=0$ , then $b = - 7$ and $c = - 2$ . The minus signs belong to the coefficients.

The quadratic formula

The quadratic formula solves any quadratic equation in standard form, even when factorising is awkward or impossible.

Definition

Quadratic formula

For $ax^2+bx+c=0$ , the solutions are

x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

The symbol $±\pm$ means “plus or minus”, so it usually gives two answers.

The expression under the square root, $b^2-4ac$ , is called the discriminant. It helps decide what kind of answers you get.

If it is positive, there are two real roots.
If it is zero, there is one repeated root.
If it is negative, there are no real roots.

Common Mistake

When the formula does not apply

The quadratic formula is not for linear equations. If $a = 0$ , there is no $x^2$ term, so the equation is not quadratic.

Example

Solving a quadratic to 2 decimal places

Solve $3x^2+4x-5=0$ , giving your answers correct to 2 decimal places.

Identify the coefficients: $a = 3$ , $b = 4$ and $c = - 5$ .
Substitute into the formula.
$x=−4±42−4(3)(−5)2(3)x=\frac{-4\pm\sqrt{4^2-4(3)(-5)}}{2(3)}$
Simplify the discriminant.
$4^2-4(3)(-5)=16+60=76$
Write the exact formula result.
$x=−4±766x=\frac{-4\pm\sqrt{76}}{6}$
Work out both values.
$x=−4+766≈0.786x=−4−766≈−2.120\begin{aligned} x&=\frac{-4+\sqrt{76}}{6}\approx 0.786\\ x&=\frac{-4-\sqrt{76}}{6}\approx -2.120 \end{aligned}$
Round to 2 decimal places: $x = 0.79$ or $x = - 2.12$ .

Tip

Calculator brackets

Type the numerator in brackets, especially for the negative answer: (-4 - sqrt(76)) ÷ 6. This avoids accidentally dividing only part of the expression.

Rounding your answers

A question may ask for answers correct to a certain number of decimal places or significant figures.

Decimal places count digits after the decimal point.
Significant figures count from the first non-zero digit.

For example, 1.6489 to 3 significant figures is 1.65, while -0.8489 to 3 significant figures is -0.849.

Example

Rearranging first, then rounding to 3 significant figures

Solve $5x^2=4x+7$ , giving your answers correct to 3 significant figures.

Rearrange into standard form.
$5x^2-4x-7=0$
Identify the coefficients: $a = 5$ , $b = - 4$ and $c = - 7$ .
Substitute into the quadratic formula.
$x=−(−4)±(−4)2−4(5)(−7)2(5)x=\frac{-(-4)\pm\sqrt{(-4)^2-4(5)(-7)}}{2(5)}$
Simplify carefully.
$x=4±16+14010x=\frac{4\pm\sqrt{16+140}}{10}$
Work out both values.
$x=4+15610≈1.649x=4−15610≈−0.849\begin{aligned} x&=\frac{4+\sqrt{156}}{10}\approx 1.649\\ x&=\frac{4-\sqrt{156}}{10}\approx -0.849 \end{aligned}$
To 3 significant figures, $x = 1.65$ or $x = - 0.849$ .

Common Mistake

Rounding too early

Do not round the square root first unless the question asks you to. Keep the full calculator value until the final answer.

Exact answers using surds

Sometimes the question does not want a decimal. It may ask for an exact answer in a form like $a±bca\pm b\sqrt{c}$ .

Definition

Surd

A surd is an irrational root left in exact form, such as $2\sqrt{2}$ or $7\sqrt{7}$ . You should simplify it if it has a square factor, for example $72=62\sqrt{72}=6\sqrt{2}$ .

Example

Giving answers in exact surd form

Solve $x^2+10x+7=0$ , giving your answers in the form $a±bca\pm b\sqrt{c}$ .

Identify $a = 1$ , $b = 10$ and $c = 7$ .
Substitute into the formula.
$x=−10±102−4(1)(7)2(1)x=\frac{-10\pm\sqrt{10^2-4(1)(7)}}{2(1)}$
Simplify the discriminant.
$10^2-4(1)(7)=100-28=72$
Simplify the surd.
$72=36⋅2=62\sqrt{72}=\sqrt{36\cdot 2}=6\sqrt{2}$
Substitute this back into the answer.
$x=−10±622x=\frac{-10\pm 6\sqrt{2}}{2}$
Divide both terms in the numerator by 2.
$x=−5±32x=-5\pm 3\sqrt{2}$

Using the formula in geometry problems

In problem-solving questions, the hard part is often forming the quadratic equation. Once you have it, the formula is the same as before.

Key Idea

Set up first, solve second

For area or triangle problems, write an equation using the geometry first. Then expand, simplify to standard form, and solve.

The diagram below shows a typical compound shape made from two rectangles. To find the total area, add the area of each rectangle.

Compound rectilinear shape made from two rectangles with algebraic side lengths

Example

Area of a compound shape

Write the area of each rectangle.
$(x - 2) (2 x + 3) + 5 (2 x - 4) = 46$
Expand each part.
$2x^2-x-6+10x-20=46$
Simplify and rearrange to standard form.
$2x^2+9x-72=0$
Use the quadratic formula with $a = 2$ , $b = 9$ and $c = - 72$ .
$x=−9±92−4(2)(−72)2(2)x=\frac{-9\pm\sqrt{9^2-4(2)(-72)}}{2(2)}$
Simplify.
$x=−9±6574x=\frac{-9\pm\sqrt{657}}{4}$
Work out both possible values: $x≈4.158x\approx 4.158$ or $x≈−8.658x\approx -8.658$ .
Reject the negative value because lengths such as $x - 2$ and $2 x - 4$ must be positive. So $x = 4.16$ to 3 significant figures.

Tip

Check physical meaning

In geometry, a negative algebraic solution may be mathematically correct but impossible for a length. Always check whether your answer makes the side lengths positive.

Right-angled triangle problems

For a right-angled triangle, the hypotenuse is the longest side, opposite the right angle. Pythagoras’ theorem says that

a^2+b^2=c^2

where $c$ is the hypotenuse.

Example

Using Pythagoras to form a quadratic

A right-angled triangle has shorter sides $x + 3$ and $x + 4$ , and hypotenuse $x + 6$ . Find the exact value of $x$ .

Use Pythagoras’ theorem.
$x+3)^2+(x+4)^2=(x+6)^2$
Expand both sides.
$x^2+6x+9+x^2+8x+16=x^2+12x+36$
Simplify and rearrange.
$x^2+2x-11=0$
Apply the quadratic formula.
$x=−2±22−4(1)(−11)2x=\frac{-2\pm\sqrt{2^2-4(1)(-11)}}{2}$
Simplify the surd.
$x=−2±482=−1±23x=\frac{-2\pm\sqrt{48}}{2}=-1\pm 2\sqrt{3}$
Only $x=−1+23x=-1+2\sqrt{3}$ is valid, because the other value would make some side lengths negative.

Exam technique

In the exam

Put the equation into $ax^2+bx+c=0$ before using the formula.
Write down $a$ , $b$ and $c$ , including negative signs.
Use brackets in your calculator, keep full accuracy until the end, and answer in the requested form: decimal, significant figures, or surd.
In geometry questions, reject any value that makes a length negative.

Self review

Check yourself

Can you rearrange an equation like $6x^2=5x+8$ into standard form and identify $a$ , $b$ and $c$ ?
When $b$ is negative, can you correctly work out $- b$ in the formula?
If an exact answer is requested, can you simplify a surd such as $80\sqrt{80}$ ?

Quadratic Formula

What you'll learn

Before the formula: standard form

The quadratic formula

Rounding your answers

Exact answers using surds

Using the formula in geometry problems

Right-angled triangle problems

Recap questions

You've reached the end

Quadratic Formula

What you'll learn

Before the formula: standard form

The quadratic formula

Rounding your answers

Exact answers using surds

Using the formula in geometry problems

Right-angled triangle problems

Recap questions

You've reached the end

Quadratic Formula

What you'll learn

Before the formula: standard form

The quadratic formula

Rounding your answers

Exact answers using surds

Using the formula in geometry problems

Right-angled triangle problems

Recap questions

You've reached the end

Quadratic Formula Revision Guide

Quadratic Formula

What you'll learn

Before the formula: standard form

The quadratic formula

Rounding your answers

Exact answers using surds

Using the formula in geometry problems

Right-angled triangle problems

Recap questions

You've reached the end

Quadratic Formula Revision Guide