1.20 Other Types of Equations

Let $A (x) = 0$ represent an equation containing $x$ that is satisfied when $x = r_{1}, r_{2}, \dots, r_{p}$ .

Let $B (x)$ be an expression in $x$ that vanishes when $x = s_{1}, s_{2}, \dots, s_{q}$ .

Then the equation

$A (x) \cdot B (x) = 0$

is satisfied not only when $x = r_{1}, r_{2}, \dots, r_{p}$ but also when $x = s_{1}, s_{2}, \dots, s_{q}$ .

So in general, when both sides of an equation in $x$ are multiplied by an expression in $x$ , the resulting equation may have solutions which the original equation did not have. The solutions that have been introduced in the process of solving an equation but do not satisfy the original equation are called extraneous solutions. In the above, $x = s_{1}, s_{2}, \dots, s_{q}$ are the extraneous solutions to the equation $A (x) = 0$ . To find the extraneous solutions, we must substitute the solutions of the resulting equation in the original equation.

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Equations involving fractional expressions

To solve equations involving fractional expressions, eliminate the denominators, for example, by multiplying each side by the least common multiple (LCM) of all denominators — although any common multiple works. Then test all the solutions to find the extraneous ones.

Example

Solve the equation $\frac{5}{x} + \frac{1}{x - 2} = \frac{4}{x^{2} - 4}$ .

Solution

To eliminate the denominators, we multiply each side by the least common multiple of all denominators: $x, x - 2, x^{2} - 4$ . Because $x^{2} - 4 = (x - 2) (x + 2)$ , the LCM of the denominators is $x (x - 2) (x + 2)$ or $x (x^{2} - 4)$

$\begin{array}{r} \begin{array}{rclr} x (x - 2) (x + 2) (\frac{5}{x} + \frac{1}{x - 2}) & = & x (x - 2) (x + 2) \frac{4}{x^{2} - 4} \\ 5 (x^{2} - 4) + x (x + 2) & = & 4 x \\ 5 x^{2} - 20 + x^{2} + 2 x & = & 4 x & (expand LHS) \\ 6 x^{2} - 2 x - 20 & = & 0 & (simplify) \end{array} \end{array}$

Using the quadratic formula, we find the solutions of the above equation:

$x = \frac{2 \pm \sqrt{4 + 240}}{12} = \frac{2 \pm 22}{12}$

so $x = - 5 / 3$ or $x = 2$ . Now we need to check if these values satisfy the original equations:

Checking

x = - 5 / 3

$\begin{aligned} Left Hand Side (LHS) & = \frac{5}{- 5 / 3} + \frac{1}{- 5 / 3 - 2} \\ = - 3 - \frac{1}{11 / 3} \\ = - 3 - \frac{3}{11} = \frac{- 36}{11} \end{aligned}$
$\begin{aligned} Right hand side (RHS) & = \frac{4}{(- 5 / 3)^{2} - 4} \\ = \frac{4}{\frac{25}{9} - 4} \\ = \frac{4}{\frac{- 11}{9}} = - \frac{36}{11} \end{aligned}$

Because LHS = RHS, $x = - 5 / 3$ is a solution.

Checking $x = 2$ :

$LHS = \frac{5}{2} + \frac{1}{2 - 2} division by zero! undefined$

$RHS = \frac{4}{4 - 4} division by zero! undefined$

so $x = 2$ is not acceptable and the only solution is $x = - 5 / 3$ .

Note that in the above example, the expression that we multiplied both sides to does not vanish when $x = - 5 / 3$ . So $x = - 5 / 3$ could not be an extraneous solution and we did not have to test whether or not it satisfies the original equation.

Radical equations

Radical equations are equations in which the variable (the unknown) is under the square root, cubic root, or higher root, like

$\sqrt{5 - 3 x} = 6, \sqrt{2 x + 7} + \sqrt{x + 3} = 10, \sqrt[3]{5 x - 7} = \sqrt[3]{4 x + 1}$

To solve radical equations:

Isolate the most complicated radical term on one side and move the rest of terms to the other side.
If the radical is a square root, square both side. If the radical is a cubic root, cube both sides and in general, for an $n$ th root radical, raise both sides to the $n$ th power. Then simplify the equation.
Repeat Steps 1 and 2 with the effort to eliminate all radicals involving the unknown.
Solve the resulting equation.
Test each solution by substitution in the original equation and determine which solutions satisfy the original equation.

Raising both sides of an equation to an even power may introduce extraneous solutions.
Recall that if $A$ is positive, $\sqrt{A}$ represents only the positive root of $A$ . Similarly $\sqrt[n]{A}$ where $n$ is even represents only the positive $n$ th root.

Example

Solve each equation:

(a) $2 + \sqrt{2 x^{2} + 4} = 2 x$

(b) $2 - \sqrt{2 x^{2} + 4} = - 2 x$

Solution

(a) Rewrite the equation as

$\sqrt{2 x^{2} + 4} = 2 x - 2$

Then we square both sides:

$\begin{array}{r} \begin{array}{rclr} 2 x^{2} + 4 & = & (2 x - 2)^{2} \\ 2 x^{2} + 4 & = & 4 x^{2} - 8 x + 4 \\ 0 & = & 2 x^{2} - 8 x & (simplify) \\ 0 & = & 2 x (x - 4) & (factor) \end{array} \end{array}$

Therefore $x = 0$ or $x = 4$ . Substituting $x = 0$ in the original equation, we get

$\begin{aligned} LHS & = 2 + \sqrt{4} = 4 \\ RHS & = 0 \end{aligned}$

Because LHS $\neq$ RHS, $x = 0$ is an extraneous solution.

Substituting $x = 4$ in the original equation, we get

$\begin{aligned} LHS & = 2 + \sqrt{2 \cdot 4^{2} + 4} = 2 + 6 = 8 \\ RHS & = 2 \cdot 4 = 8 \end{aligned}$

and LHS $=$ RHS. So the only solution of this equation is $x = 4$ .

(b) Rewrite the equation as

$- \sqrt{2 x^{2} + 4} = - (2 + 2 x)$

$\begin{array}{r} \begin{array}{rclr} \sqrt{2 x^{2} + 4} & = 2 + 2 x \\ 2 x^{2} + 4 & = & (2 + 2 x)^{2} & (square both sides) \\ 2 x^{2} + 4 & = & 4 + 8 x + 4 x^{2} & (expand) \\ 0 & = & 2 x^{2} + 8 x & (simplify) \\ 0 & = & 2 x (x + 4) \end{array} \end{array}$

which gives $x = 0$ or $x = - 4$ . Substituting $x = 0$ in the original equation, we get

$\begin{aligned} LHS & = 2 - \sqrt{4} = 0, \\ RHS & = - 2 \cdot 0 = 0 \end{aligned}$

Because LHS $=$ RHS, $x = 0$ is a solution. Substituting $x = - 4$ in the original equation, we get

$\begin{aligned} LHS & = 2 - \sqrt{2 \cdot (- 4)^{2} + 4} = 2 - 6 = - 4 \\ RHS & = - 2 \cdot (- 4) = 8 \end{aligned}$

Because LHS $\neq$ RHS, $x = - 4$ is not a solution (it is an extraneous solution) and so $x = 0$ is the only solution.

Example

Solve each equation:

(a) $\sqrt{x + 19} + \sqrt{x + 10} = 9.$

(b) $\sqrt{x - 2} - \sqrt{2 x + 5} = 3.$

Solution

(a)

$\begin{array}{r} \begin{array}{rcll} \sqrt{x + 19} & = & 9 - \sqrt{x + 10} & (rewrite the equation) \\ x + 19 & = & 81 - 18 \sqrt{x + 10} + x + 10 & (square both sides) \\ 18 \sqrt{x + 10} & = & 72 & (simplify) \\ \sqrt{x + 10} & = & 4 \\ x + 10 & = & 16 & (square both sides) \\ x & = 6 \end{array} \end{array}$

Let’s check if $x = 6$ satisfies the original equation:

$\begin{aligned} LHS & = \sqrt{6 + 19} + \sqrt{6 + 10} = \sqrt{25} + \sqrt{16} = 5 + 4 \\ RHS & = 9 \end{aligned}$

Because LHS $=$ RHS, $x = 6$ is the solution.

(b)

$\begin{array}{r} \begin{array}{rclr} \sqrt{x - 2} & = & 3 + \sqrt{2 x + 5} & (rewrite the equation) \\ x - 2 & = & 9 + 6 \sqrt{2 x + 6} + 2 x + 5 & (square both sides) \\ - 6 \sqrt{2 x + 6} & = & x + 16 & (simplify) \\ 36 (2 x + 6) & = & x^{2} + 32 x + 16 & (square both sides) \\ x^{2} - 40 x + 76 & = & 0 & (simplify) \end{array} \end{array}$

Now we can factor it by trial and error as

$x^{2} - 40 x + 76 = (x - 2) (x - 38)$

which gives $x = 2$ or $x = 38$ . Alternatively we can use the quadratic equation to find the roots

$\begin{aligned} x & = \frac{40 \pm \sqrt{1600 - 4 \cdot 76}}{2} \\ = \frac{40 \pm \sqrt{1296}}{2} \\ = \frac{40 \pm 36}{2} \\ = 20 \pm 18 \end{aligned}$

which gives $x = 2$ or $x = 38$ . We have to test these values to see if they satisfy the original equation.

Substituting 2 for $x$ :

$\begin{aligned} LHS & = \sqrt{2 - 2} - \sqrt{2 \cdot 2 + 5} = - 3, \\ RHS & = 3. \end{aligned}$

Because LHS $\neq$ RHS, $x = 2$ is not a solution.

Substituting 38 for $x$ :

$\begin{aligned} LHS & = \sqrt{38 - 2} - \sqrt{2 \cdot 38 + 5} = 6 - 9 = - 3, \\ RHS & = 3. \end{aligned}$

Because LHS $\neq$ RHS, $x = 38$ is not a solution. Therefore, this problem does not have any solutions.