Quadratic Formula

What you'll learn

How to put a quadratic equation into standard form.
How to substitute into the quadratic formula without losing signs.
How to give answers as decimals or exact surds.
How quadratic formula questions can appear in shape and Pythagoras problems.

Why we need the quadratic formula

You already know some quadratics can be solved by factorising, such as finding two brackets. But not every quadratic factorises neatly.

The quadratic formula is useful because it works for any quadratic equation in standard form, including ones with awkward decimal or surd answers.

Definition

Quadratic in standard form

A quadratic equation is an equation where the highest power of the variable is squared. In standard form, it is

Standard form separates the squared term, linear term and constant so the coefficients a, b and c can be identified.

ax^2+bx+c=0

where $a$ , $b$ and $c$ are coefficients: the numbers multiplying the terms. Also, $\neq 0$ .

Step 1: Rearrange into standard form

Before using the formula, everything must be on one side and the other side must be zero.

Key Idea

Standard form first

Do not start substituting until your equation looks like $ax^2+bx+c=0$ . The signs of $b$ and $c$ often change when you rearrange.

Example

Rearranging before solving

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

Moving every term to the left changes the signs of the 4x and 7 terms before using the formula.

Move everything to the left-hand side by subtracting $4 x$ and subtracting 7:
$5x^2-4x-7=0$
Identify the coefficients:
$a=5,b=−4,c=−7a=5,\quad b=-4,\quad c=-7$
Substitute into the formula:
$x=−(−4)±(−4)2−4(5)(−7)2(5)x=\frac{-(-4)\pm\sqrt{(-4)^2-4(5)(-7)}}{2(5)}$
Simplify carefully:
$x=4±15610x=\frac{4\pm\sqrt{156}}{10}$
Use your calculator to find both answers:
$x=1.648…orx=−0.848…x=1.648\ldots \quad \text{or} \quad x=-0.848\ldots$
Round to 3 significant figures:
$\quad \text{or} \quad x=-0.848$

Common Mistake

Forgetting to rearrange

If the equation is written like $5x^2=4x+7$ , then $b$ is not 4 and $c$ is not 7. You must first write it as $5x^2-4x-7=0$ .

Step 2: Use the quadratic formula

Definition

The quadratic formula

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

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

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

The plus-minus symbol means the formula splits into a plus calculation and a minus calculation.

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

Tip

Two calculator entries

After finding the square root part, do the “plus” answer and the “minus” answer separately. This helps avoid typing the whole formula incorrectly.

Example

Solving to decimal places

Solve $2x^2+9x+3=0$ , giving your answers correct to 2 decimal places.

The quadratic has two x-intercepts, matching the two solutions found using the formula.

Identify the coefficients:
$a=2,b=9,c=3a=2,\quad b=9,\quad c=3$
Substitute into the quadratic formula:
$x=−9±92−4(2)(3)2(2)x=\frac{-9\pm\sqrt{9^2-4(2)(3)}}{2(2)}$
Simplify inside the square root and the denominator:
$x=−9±574x=\frac{-9\pm\sqrt{57}}{4}$
Work out the “plus” solution:
$x=−9+574=−0.362…x=\frac{-9+\sqrt{57}}{4}=-0.362\ldots$
Work out the “minus” solution:
$x=−9−574=−4.137…x=\frac{-9-\sqrt{57}}{4}=-4.137\ldots$
Round both answers to 2 decimal places:
$\quad \text{or} \quad x=-4.14$

Rounding: decimal places and significant figures

A common instruction is “give your solutions correct to 2 decimal places” or “3 significant figures”.

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

For example, 1.648 to 3 significant figures is 1.65, while -0.8482 to 3 significant figures is -0.848.

Common Mistake

Rounding too early

Do not round the square root halfway through. Keep the full calculator value until the final line, then round your two answers.

Example

Solving to 3 significant figures

Solve $4x^2+x-9=0$ , giving your answers correct to 3 significant figures.

Identify the coefficients:
$a=4,b=1,c=−9a=4,\quad b=1,\quad c=-9$
Substitute:
$x=−1±12−4(4)(−9)2(4)x=\frac{-1\pm\sqrt{1^2-4(4)(-9)}}{2(4)}$
Simplify:
$x=−1±1458x=\frac{-1\pm\sqrt{145}}{8}$
Find the two calculator values:
$x=1.380…orx=−1.630…x=1.380\ldots \quad \text{or} \quad x=-1.630\ldots$
Round to 3 significant figures:
$\quad \text{or} \quad x=-1.63$

Exact answers in surd form

Sometimes you are asked to give answers in a form like $a±bca\pm b\sqrt{c}$ instead of decimals.

Definition

Surd

A surd is an exact square root that cannot be simplified to a whole number, such as $3\sqrt{3}$ or $7\sqrt{7}$ .

To give exact answers, use the quadratic formula but do not press the decimal button. Instead, simplify the square root where possible.

Example

Giving exact surd answers

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

The exact surd answers can be represented as two exact positions on the x-axis rather than rounded decimals.

Identify the coefficients:
$a=1,b=10,c=−2a=1,\quad b=10,\quad c=-2$
Substitute into the formula:
$x=−10±102−4(1)(−2)2(1)x=\frac{-10\pm\sqrt{10^2-4(1)(-2)}}{2(1)}$
Simplify inside the square root:
$x=−10±1082x=\frac{-10\pm\sqrt{108}}{2}$
Simplify the surd:
$108=36⋅3=63\sqrt{108}=\sqrt{36\cdot 3}=6\sqrt{3}$
Divide both parts of the numerator by 2:
$x=−5±33x=-5\pm3\sqrt{3}$

Tip

Simplifying surds

Look for the biggest square number that divides the number under the root. For example, 108 contains 36, so $108=63\sqrt{108}=6\sqrt{3}$ .

Quadratics from shape problems

Some exam questions ask you to form a quadratic from an area or Pythagoras problem, then solve it.

For area problems, remember:

Area of a rectangle = length multiplied by width.
Total area = area of each part added together.

Example

Forming and solving a quadratic from area

A shape is made from two rectangles. One rectangle has width $x - 2$ and height $2 x + 3$ . The other rectangle has width $x + 4$ and height 5. The total area is 40 cm². Find $x$ to 3 significant figures.

The total area is the sum of the areas of the two rectangles.

Write an expression for the total area:
$(x - 2) (2 x + 3) + 5 (x + 4) = 40$
Expand the brackets:
$2x^2-x-6+5x+20=40$
Collect like terms:
$2x^2+4x+14=40$
Rearrange into standard form:
$2x^2+4x-26=0$
Divide by 2 to make it simpler:
$x^2+2x-13=0$
Use the quadratic formula with $a = 1$ , $b = 2$ and $c = - 13$ :
$x=−2±22−4(1)(−13)2x=\frac{-2\pm\sqrt{2^2-4(1)(-13)}}{2}$
Simplify:
$x=−1±14x=-1\pm\sqrt{14}$
Choose the positive value because lengths cannot be negative:
$x=2.741…x=2.741\ldots$
Round to 3 significant figures:
$x = 2.74$

Common Mistake

Check the context

In geometry problems, a negative solution may be algebraically correct but impossible as a length. Always check which root actually fits the question.

Quadratics from Pythagoras

For a right-angled triangle, use Pythagoras’ theorem:

a^2+b^2=c^2

where $c$ is the hypotenuse, the longest side opposite the right angle.

In a right-angled triangle, the hypotenuse is opposite the right angle and is the side used as c in Pythagoras’ theorem.

Example

Using Pythagoras to form a quadratic

A right-angled triangle has shorter sides $x + 1$ and $x + 4$ , and hypotenuse $x + 6$ . Find the valid value of $x$ exactly.

The hypotenuse is the longest side, so it goes on the right-hand side of Pythagoras’ theorem.

Use Pythagoras’ theorem:
$x+1)^2+(x+4)^2=(x+6)^2$
Expand each bracket:
$x^2+2x+1+x^2+8x+16=x^2+12x+36$
Collect terms and rearrange:
$x^2-2x-19=0$
Use the quadratic formula:
$x=2±(−2)2−4(1)(−19)2x=\frac{2\pm\sqrt{(-2)^2-4(1)(-19)}}{2}$
Simplify:
$x=2±802x=\frac{2\pm\sqrt{80}}{2}$
Simplify the surd:
$x=1±25x=1\pm2\sqrt{5}$
Choose the positive value because the side lengths must be positive:
$x=1+25x=1+2\sqrt{5}$

Exam technique

In the exam

Rearrange to $ax^2+bx+c=0$ first, then label $a$ , $b$ and $c$ including their signs.
Write a clear substitution line before using your calculator, because method marks are often available.
Follow the answer instruction exactly: decimal places, significant figures, or exact surd form.
In shape problems, reject any answer that would make a length negative or impossible.

Self review

Check yourself

Can you identify $a$ , $b$ and $c$ after rearranging $3x^2=2x+8$ ?
What does the $±\pm$ symbol mean in the quadratic formula?
Why might a quadratic from a geometry question have two algebraic answers but only one valid value of $x$ ?

What you'll learn

How to put a quadratic equation into standard form.
How to substitute into the quadratic formula without losing signs.
How to give answers as decimals or exact surds.
How quadratic formula questions can appear in shape and Pythagoras problems.

Why we need the quadratic formula

You already know some quadratics can be solved by factorising, such as finding two brackets. But not every quadratic factorises neatly.

The quadratic formula is useful because it works for any quadratic equation in standard form, including ones with awkward decimal or surd answers.

Definition

Quadratic in standard form

A quadratic equation is an equation where the highest power of the variable is squared. In standard form, it is

Standard form separates the squared term, linear term and constant so the coefficients a, b and c can be identified.

ax^2+bx+c=0

where $a$ , $b$ and $c$ are coefficients: the numbers multiplying the terms. Also, $\neq 0$ .

Step 1: Rearrange into standard form

Before using the formula, everything must be on one side and the other side must be zero.

Key Idea

Standard form first

Do not start substituting until your equation looks like $ax^2+bx+c=0$ . The signs of $b$ and $c$ often change when you rearrange.

Example

Rearranging before solving

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

Moving every term to the left changes the signs of the 4x and 7 terms before using the formula.

Move everything to the left-hand side by subtracting $4 x$ and subtracting 7:
$5x^2-4x-7=0$
Identify the coefficients:
$a=5,b=−4,c=−7a=5,\quad b=-4,\quad c=-7$
Substitute into the formula:
$x=−(−4)±(−4)2−4(5)(−7)2(5)x=\frac{-(-4)\pm\sqrt{(-4)^2-4(5)(-7)}}{2(5)}$
Simplify carefully:
$x=4±15610x=\frac{4\pm\sqrt{156}}{10}$
Use your calculator to find both answers:
$x=1.648…orx=−0.848…x=1.648\ldots \quad \text{or} \quad x=-0.848\ldots$
Round to 3 significant figures:
$\quad \text{or} \quad x=-0.848$

Common Mistake

Forgetting to rearrange

If the equation is written like $5x^2=4x+7$ , then $b$ is not 4 and $c$ is not 7. You must first write it as $5x^2-4x-7=0$ .

Step 2: Use the quadratic formula

Definition

The quadratic formula

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

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

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

The plus-minus symbol means the formula splits into a plus calculation and a minus calculation.

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

Tip

Two calculator entries

After finding the square root part, do the “plus” answer and the “minus” answer separately. This helps avoid typing the whole formula incorrectly.

Example

Solving to decimal places

Solve $2x^2+9x+3=0$ , giving your answers correct to 2 decimal places.

The quadratic has two x-intercepts, matching the two solutions found using the formula.

Identify the coefficients:
$a=2,b=9,c=3a=2,\quad b=9,\quad c=3$
Substitute into the quadratic formula:
$x=−9±92−4(2)(3)2(2)x=\frac{-9\pm\sqrt{9^2-4(2)(3)}}{2(2)}$
Simplify inside the square root and the denominator:
$x=−9±574x=\frac{-9\pm\sqrt{57}}{4}$
Work out the “plus” solution:
$x=−9+574=−0.362…x=\frac{-9+\sqrt{57}}{4}=-0.362\ldots$
Work out the “minus” solution:
$x=−9−574=−4.137…x=\frac{-9-\sqrt{57}}{4}=-4.137\ldots$
Round both answers to 2 decimal places:
$\quad \text{or} \quad x=-4.14$

Rounding: decimal places and significant figures

A common instruction is “give your solutions correct to 2 decimal places” or “3 significant figures”.

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

For example, 1.648 to 3 significant figures is 1.65, while -0.8482 to 3 significant figures is -0.848.

Common Mistake

Rounding too early

Do not round the square root halfway through. Keep the full calculator value until the final line, then round your two answers.

Example

Solving to 3 significant figures

Solve $4x^2+x-9=0$ , giving your answers correct to 3 significant figures.

Identify the coefficients:
$a=4,b=1,c=−9a=4,\quad b=1,\quad c=-9$
Substitute:
$x=−1±12−4(4)(−9)2(4)x=\frac{-1\pm\sqrt{1^2-4(4)(-9)}}{2(4)}$
Simplify:
$x=−1±1458x=\frac{-1\pm\sqrt{145}}{8}$
Find the two calculator values:
$x=1.380…orx=−1.630…x=1.380\ldots \quad \text{or} \quad x=-1.630\ldots$
Round to 3 significant figures:
$\quad \text{or} \quad x=-1.63$

Exact answers in surd form

Sometimes you are asked to give answers in a form like $a±bca\pm b\sqrt{c}$ instead of decimals.

Definition

Surd

A surd is an exact square root that cannot be simplified to a whole number, such as $3\sqrt{3}$ or $7\sqrt{7}$ .

To give exact answers, use the quadratic formula but do not press the decimal button. Instead, simplify the square root where possible.

Example

Giving exact surd answers

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

The exact surd answers can be represented as two exact positions on the x-axis rather than rounded decimals.

Identify the coefficients:
$a=1,b=10,c=−2a=1,\quad b=10,\quad c=-2$
Substitute into the formula:
$x=−10±102−4(1)(−2)2(1)x=\frac{-10\pm\sqrt{10^2-4(1)(-2)}}{2(1)}$
Simplify inside the square root:
$x=−10±1082x=\frac{-10\pm\sqrt{108}}{2}$
Simplify the surd:
$108=36⋅3=63\sqrt{108}=\sqrt{36\cdot 3}=6\sqrt{3}$
Divide both parts of the numerator by 2:
$x=−5±33x=-5\pm3\sqrt{3}$

Tip

Simplifying surds

Look for the biggest square number that divides the number under the root. For example, 108 contains 36, so $108=63\sqrt{108}=6\sqrt{3}$ .

Quadratics from shape problems

Some exam questions ask you to form a quadratic from an area or Pythagoras problem, then solve it.

For area problems, remember:

Area of a rectangle = length multiplied by width.
Total area = area of each part added together.

Example

Forming and solving a quadratic from area

The total area is the sum of the areas of the two rectangles.

Write an expression for the total area:
$(x - 2) (2 x + 3) + 5 (x + 4) = 40$
Expand the brackets:
$2x^2-x-6+5x+20=40$
Collect like terms:
$2x^2+4x+14=40$
Rearrange into standard form:
$2x^2+4x-26=0$
Divide by 2 to make it simpler:
$x^2+2x-13=0$
Use the quadratic formula with $a = 1$ , $b = 2$ and $c = - 13$ :
$x=−2±22−4(1)(−13)2x=\frac{-2\pm\sqrt{2^2-4(1)(-13)}}{2}$
Simplify:
$x=−1±14x=-1\pm\sqrt{14}$
Choose the positive value because lengths cannot be negative:
$x=2.741…x=2.741\ldots$
Round to 3 significant figures:
$x = 2.74$

Common Mistake

Check the context

In geometry problems, a negative solution may be algebraically correct but impossible as a length. Always check which root actually fits the question.

Quadratics from Pythagoras

For a right-angled triangle, use Pythagoras’ theorem:

a^2+b^2=c^2

where $c$ is the hypotenuse, the longest side opposite the right angle.

In a right-angled triangle, the hypotenuse is opposite the right angle and is the side used as c in Pythagoras’ theorem.

Example

Using Pythagoras to form a quadratic

A right-angled triangle has shorter sides $x + 1$ and $x + 4$ , and hypotenuse $x + 6$ . Find the valid value of $x$ exactly.

The hypotenuse is the longest side, so it goes on the right-hand side of Pythagoras’ theorem.

Use Pythagoras’ theorem:
$x+1)^2+(x+4)^2=(x+6)^2$
Expand each bracket:
$x^2+2x+1+x^2+8x+16=x^2+12x+36$
Collect terms and rearrange:
$x^2-2x-19=0$
Use the quadratic formula:
$x=2±(−2)2−4(1)(−19)2x=\frac{2\pm\sqrt{(-2)^2-4(1)(-19)}}{2}$
Simplify:
$x=2±802x=\frac{2\pm\sqrt{80}}{2}$
Simplify the surd:
$x=1±25x=1\pm2\sqrt{5}$
Choose the positive value because the side lengths must be positive:
$x=1+25x=1+2\sqrt{5}$

Exam technique

In the exam

Rearrange to $ax^2+bx+c=0$ first, then label $a$ , $b$ and $c$ including their signs.
Write a clear substitution line before using your calculator, because method marks are often available.
Follow the answer instruction exactly: decimal places, significant figures, or exact surd form.
In shape problems, reject any answer that would make a length negative or impossible.

Self review

Check yourself

Can you identify $a$ , $b$ and $c$ after rearranging $3x^2=2x+8$ ?
What does the $±\pm$ symbol mean in the quadratic formula?
Why might a quadratic from a geometry question have two algebraic answers but only one valid value of $x$ ?

What you'll learn

Why we need the quadratic formula

Step 1: Rearrange into standard form

Step 2: Use the quadratic formula

Rounding: decimal places and significant figures

Exact answers in surd form

Quadratics from shape problems

Quadratics from Pythagoras

Recap questions

Review Quadratic Formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Quadratic Formula

What you'll learn

Why we need the quadratic formula

Step 1: Rearrange into standard form

Step 2: Use the quadratic formula

Rounding: decimal places and significant figures

Exact answers in surd form

Quadratics from shape problems

Quadratics from Pythagoras

Recap questions

Review Quadratic Formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Quadratic Formula

What you'll learn

Why we need the quadratic formula

Step 1: Rearrange into standard form

Step 2: Use the quadratic formula

Rounding: decimal places and significant figures

Exact answers in surd form

Quadratics from shape problems

Quadratics from Pythagoras

Recap questions

Review Quadratic Formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Quadratic Formula Revision Guide

Quadratic Formula

What you'll learn

Why we need the quadratic formula

Step 1: Rearrange into standard form

Step 2: Use the quadratic formula

Rounding: decimal places and significant figures

Exact answers in surd form

Quadratics from shape problems

Quadratics from Pythagoras

Recap questions

Review Quadratic Formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Quadratic Formula Revision Guide