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# 2.4 Substitution

## 2.4 Substitution (EMBFM)

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.

## Worked example 10: Solving by substitution

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

$k - \frac{3}{k} = 2$

### 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 - \frac{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 $$x = 3$$.

temp text
Textbook Exercise 2.4

Solve the following quadratic equations by substitution:

$$-24 = 10(x^2 + 5x) + (x^2 + 5x)^2$$

We notice that $$x^2 + 5x$$ is a repeated expression and we therefore let $$k = x^2 + 5x$$ so that the equation becomes

$-24 = 10k + k^{2}$

Now we can solve for $$k$$:

\begin{aligned} k^{2} + 10k + 24 &= 0 \\ (k + 6)(k + 4) &= 0 \\ \text{Therefore } k = -6 &\text{ or } k = -4 \end{aligned}

We now use these values for $$k$$ to solve for the original variable $$x$$

For $$k = -6$$ \begin{aligned} x^2 + 5x &= -6 \\ x^2 + 5x + 6 &= 0 \\ (x + 2)(x + 3) &= 0 \\ \text{Therefore } x = -2 & \text{ or } x = -3 \end{aligned}

For $$k = -4$$ \begin{aligned} x^2 + 5x &= -4 \\ x^2 + 5x + 4 &= 0 \\ (x + 1)(x + 4) &= 0 \\ \text{Therefore } x = -1 & \text{ or } x = -4 \end{aligned}

The roots of the equation are $$x = -1$$, $$x = -4$$, $$x = -2$$ and $$x = -3$$.

$$(x^2 - 2x)^2 - 8 = 7(x^2 - 2x)$$

We notice that $$x^2 - 2x$$ is a repeated expression and we therefore let $$k = x^2 - 2x$$ so that the equation becomes

$k^{2} - 8 = 7k$

Now we can solve for $$k$$:

\begin{aligned} k^{2} - 7k - 8 &= 0 \\ (k + 1)(k - 8) &= 0 \\ \text{Therefore } k = -1 &\text{ or } k = 8 \end{aligned}

We now use these values 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 $$k = 8$$ \begin{aligned} x^2 - 2x &= 8 \\ x^2 - 2x - 8 &= 0 \\ (x + 2)(x - 4) &= 0 \\ \text{Therefore } x = -2 & \text{ or } x = 4 \end{aligned}

The roots of the equation are $$x = 1$$, $$x = 4$$ and $$x = -2$$.

$$x^2 + 3x - \dfrac{56}{x(x + 3)} = 26$$

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 $$x \ne -3$$.

We note that the denominator is equivalent to $$x^2 + 3x$$.

We notice that $$x^2 + 3x$$ is a repeated expression and we therefore let $$k = x^2 + 3x$$ so that the equation becomes

$k - \frac{56}{k} = 26$

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 $$k \ne 0$$.

Now we can solve for $$k$$

\begin{aligned} k - \frac{56}{k}&= 26 \\ k^{2} - 56 &= 26k \\ k^{2} - 26k - 56 &= 0 \\ (k - 28)(k + 2)&= 0 \\ \text{Therefore } k = 28 &\text{ or } k = -2 \end{aligned}

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

We can now use values obtained for $$k$$ to solve for the original variable $$x$$

For $$k = 28$$ \begin{aligned} x^2 + 3x &= 28 \\ x^2 + 3x - 28 &= 0 \\ (x + 7)(x - 4) &= 0 \\ \text{Therefore } x = -7 & \text{ or } x = 4 \end{aligned}

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

We check these roots against the restrictions for $$x$$ and confirm that all four values are valid.

The roots of the equation are $$x = -7$$, $$x = 4$$, $$x = -1$$ and $$x = -2$$.

$$x^2 - 18 + x + \dfrac{72}{x^2 + x} = 0$$

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 $$x \ne -1$$.

We notice that $$x^2 + x$$ is a repeated expression and we therefore let $$k = x^2 + x$$ so that the equation becomes

$k - 18 + \frac{72}{k} = 0$

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 $$k \ne 0$$.

Now we can solve for $$k$$

\begin{aligned} k - 18 + \frac{72}{k}&= 0\\ k^{2} - 18k + 72 &= 0 \\ (k - 12)(k - 6)&= 0 \\ \text{Therefore } k = 12 &\text{ or } k = 6 \end{aligned}

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

We can now use values obtained for $$k$$ to solve for the original variable $$x$$

For $$k = 12$$ \begin{aligned} x^2 + x &= 12 \\ x^2 + x - 12 &= 0 \\ (x + 4)(x - 3) &= 0 \\ \text{Therefore } x = -4 & \text{ or } x = 3 \end{aligned}

For $$k = 6$$ \begin{aligned} x^2 + x &= 6 \\ x^2 + x - 6 &= 0 \\ (x + 3)(x - 2)&= 0 \\ \text{Therefore } x = -3 &\text{ or } x = 2 \end{aligned}

We check these roots against the restrictions for $$x$$ and confirm that all four values are valid.

The roots of the equation are $$x = -4$$, $$x = 3$$, $$x = -3$$ and $$x = 2$$.

$$x^2 - 4x + 10 - 7(4x - x^2) = -2$$

\begin{aligned} x^{2} - 4x + 10 - 28x + 7x^{2} & = -2 \\ 8x^{2} - 32x + 12 &= 0 \\ 2x^{2} - 8x + 3 & = 0 \\ x &= \dfrac{-(-8) \pm \sqrt{(-8)^{2} - 4(2)(3)}}{2(2)}\\ &= \dfrac{8 \pm \sqrt{64 - 24}}{4}\\ &= \dfrac{8 \pm \sqrt{40}}{4}\\ \text{therefore } x =\dfrac{8 + \sqrt{40}}{4} &\text{ or } x =\dfrac{8 - \sqrt{40}}{4} \end{aligned}

$$\dfrac{9}{x^2 + 2x -12} = x^2 + 2x -12$$

To find the restrictions we first note that the roots of the denominator are $$\frac{-2 \pm \sqrt{52}}{2}$$

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 $$x \ne \frac{-2 + \sqrt{52}}{2}$$ and $$x \ne \frac{-2 - \sqrt{52}}{2}$$.

We notice that $$x^2 + 2x - 12$$ is a repeated expression and we therefore let $$k = x^2 + 2x - 12$$ so that the equation becomes

$\frac{9}{k} = k$

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 $$k \ne 0$$.

Now we can solve for $$k$$

\begin{aligned} \frac{9}{k}&= k\\ 9 &= k^{2} \\ k &= \pm 3 \\ \text{Therefore } k = -3 &\text{ or } k = 3 \end{aligned}

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

We can now use values obtained for $$k$$ to solve for the original variable $$x$$

For $$k = 3$$ \begin{aligned} x^2 + 2x - 12 &= 3 \\ x^2 + 2x - 15 &= 0 \\ (x + 5)(x - 3) &= 0 \\ \text{Therefore } x = -5 & \text{ or } x = 3 \end{aligned}

For $$k = -3$$ \begin{aligned} x^2 + 2x - 12 &= -3 \\ x^2 + 2x - 9 &= 0 \\ x &= \dfrac{-2 \pm \sqrt{(2)^{2} - 4(1)(-9)}}{2(1)}\\ &= \dfrac{-2 \pm \sqrt{4 + 36}}{2}\\ &= \dfrac{-2 \pm \sqrt{40}}{2}\\ &= \dfrac{-2 \pm 2\sqrt{10}}{2}\\ &= -1 \pm \sqrt{10}\\ \text{therefore } x = -1 + \sqrt{10} &\text{ or } x = -1 - \sqrt{10} \end{aligned}

We check these roots against the restrictions for $$x$$ and confirm that all four values are valid.

The roots of the equation are $$x = -5$$, $$x = 3$$, $$x = -1 + \sqrt{10}$$ and $$x = -1 - \sqrt{10}$$.