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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.

Write the final answer

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

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 \(x = -1 - \sqrt{10}\).