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An equation is a mathematical statement showing that two expressions are equal.

Equations can be solved by finding values (solutions) which make the statement true.

A linear equation has no terms raised to the power of 2 or higher. It has just one solution - one value that makes the equations true.

For example

\(y + 7 = 22\)

\(3m + 12 = 10 - m\)

\(3 (x - 1) = 27\)

A quadratic equation has at least one \(x^2\) term. In general, a quadratic equation has two solutions.

For example

\(x^2 = 9\)

\(3x^2 + x = 14\)

\(x^2 - 5x + 6 = 0\)

\(2 - 3x - 5x^2 = 0\)

An identity is a mathematical statement which is true for all values of a variable.

For example

\(5x + 4 = 3x + 2x + 3 + 1\) is an identity.

In this example, the right hand side (RHS) can be simplified to give the expression on the left hand side (LHS). Therefore, for any value of \(x\). the right hand side will always be the same as the left hand side.

The symbol \(\equiv\) is sometimes used to indicate an identity.

Other examples

\(3 (x - 2) = 3x - 6\)

\(x^2 - 9 = (x+3) (x-3)\)

\((x + 1) (x + 3) = x^2 + 4x + 3\)

The general form of a quadratic equation is

\(ax^2 + bx + c = 0\)

Where \(a\), \(b\) and \(c\) are numbers.

Both \(b\) and / or \(c\) can be equal to zero.

In this section, solving equations where \(a = 1\) will be considered.

To solve a quadratic equation

• rearrange the equation so that the RHS = 0
• factorise the LHS
• put the first factor equal to 0 and solve the linear equation
• put the second factor equal to 0 and solve the linear equation
• write down the two solutions

Example

Solve the equation \(x^2 - x = 5x\)

Solution

Rearrange the equation so that the RHS = 0

Subtract \(-5x\) from both sides to give 0 on RHS

Then simplify the LHS

\(x^2 - x - 5x = 5x - 5x\)

\(x^2 - 6x = 0\)

Factorise the LHS

\(x (x - 6) = 0\)

\(x\) is the common factor of \(x^2\) and \(6x\)

Put the first factor equal to 0 and solve the linear equation

\(x = 0\)

Put the second factor equal to 0 and solve the linear equation

\(x - 6 = 0\)

\(x = 6\)

Answer

Don't forget to give both solutions in your answer.

\(x = 6\) or \(0\)

Example

Solve the equation \(x^2 + 12 = 7x\)

Solution

Rearrange the equation so that the RHS = 0

\(x^2 + 12 - 7x = 0\)

Change the order of the terms to the more familiar \( x^2 - bx + c = 0\) form.

It will be easier to factorise.

\(x^2 - 7x + 12 = 0\)

Factorise the LHS

\(x^2 - 7x + 12 = 0\)

Sum = \(-7\)

Product = \(+12\)

Key numbers are \(-4\) and \(-3\)

\((x - 4) (x - 3) = 0\)

Put the first factor equal to 0 and solve the linear equation

\(x - 4 = 0\)

\(x = 4\)

Put the second factor equal to 0 and solve the linear equation

\(x - 3 = 0\)

\(x = 3\)

Answer

Write down the two solutions

\(x = 3\) or \(4\)

Example

Solve the equation \(x^2 - 16 = 0\)

Solution

No need to rearrange as RHS is already equal to zero.

Factorise the LHS - use the difference of two squares

\((x + 4) (x - 4) = 0\)

Put the first factor equal to 0 and solve the linear equation

\(x + 4 = 0\)

\(x = -4\)

Put the second factor equal to 0 and solve the linear equation

\(x - 4 = 0\)

\(x = +4\)

Answer

Write down the two solutions

\(x = -4\) or \(+4\)

Alternative solution

\(x^2 - 16 = 0\)

\(x^2 = 16\)

Take the square root of both sides

\(x = \sqrt{16}\)

Remember

\(\sqrt{x^2} = x\)

\(\sqrt{16} = + 4\) or \(-4\)

Don't forget the negative square root!

Answer

\(x = +4\) or \(-4\)

Example

The diagram shows a triangle with sides measured in centimetres. The perimeter of the triangle is 21cm.

Form an equation and solve it to find the value of \(x\)

A solution by trial and improvement will not be accepted

Solution

The perimeter is the sum of the three sides.

\(P = x + 4 + x + 3 + x^2\)

Simplify \( = x^2 + 2x + 6\)

Perimeter = \(21 \)

\(x^2 + 2x + 6 = 21\)

Rearrange to have 0 on RHS

\(x^2 + 2 x - 15 = 0\)

Solve

Product = \(-15\)

Sum = \(+2\)

Key numbers are \(+5\) and \(-3\)

\(x^2 + 2x - 15 = (x + 5) (x - 3)\)

\(x + 5 = 0\)

\(x = -5\)

\(x - 3 = 0\)

\(x = 3\)

(\(x\) cannot equal \(-5\), this would make one of the sides negative in length as \(x + 4 = -1\))

Answer

\(x = 3cm\)

A quadratic equation usually has two solutions. In these types of questions, one of the answers will make sense and the other will not. Here, \(x = 3\) is the answer.