\(-24 = 10(x^2 + 5x) + (x^2 + 5x)^2\)
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 \(x = -1 - \sqrt{10}\).
