# Substitution

## Substitution

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.

The roots of the equation are $$x = -1$$, $$x = 1$$ and $$x = 3$$.