## Substitution

Contents

It is often useful to make a substitution for a repeated expression in a quadratic equation. This makes the equation simpler and much easier to solve.

## Example

### Question

Solve for \(x\): \(x^2 – 2x – \dfrac{3}{x^2 – 2x} = 2\)

### Determine the restrictions for \(x\)

The restrictions are the values for \(x\) that would result in the denominator being equal to \(\text{0}\), which would make the fraction undefined. Therefore \(x \ne 0\) and \(x \ne 2\).

### Substitute a single variable for the repeated expression

We notice that \(x^2 – 2x\) is a repeated expression and we therefore let \(k = x^2 – 2x\) so that the equation becomes

\[k – \cfrac{3}{k} = 2\]

### Determine the restrictions for \(k\)

The restrictions are the values for \(k\) that would result in the denominator being equal to \(\text{0}\), which would make the fraction undefined. Therefore \(k \ne 0\).

### Solve for \(k\)

\[\begin{aligned} k – \cfrac{3}{k}&= 2 \\ k^2 – 3 &= 2k \\ k^2 – 2k – 3 &= 0 \\ (k + 1)(k – 3)&= 0 \\ \text{Therefore } k = -1 &\text{ or } k = 3 \end{aligned}\]

We check these two roots against the restrictions for \(k\) and confirm that both are valid.

### Use values obtained for \(k\) to solve for the original variable \(x\)

For \(k = -1\) \[\begin{aligned} x^2 – 2x &= -1 \\ x^2 – 2x + 1 &= 0 \\ (x – 1)(x – 1) &= 0 \\ \text{Therefore } x &= 1 \end{aligned}\]

For \(k = 3\) \[\begin{aligned} x^2 – 2x &= 3 \\ x^2 – 2x – 3 &= 0 \\ (x + 1)(x – 3)&= 0 \\ \text{Therefore } x = -1 &\text{ or } x = 3 \end{aligned}\]

We check these roots against the restrictions for \(x\) and confirm that all three values are valid.

### Write the final answer

The roots of the equation are \(x = -1\), \(x = 1\) and \(x = 3\).