Before you read

Answer first — see what you already know.

A pendulum formula gives $x + 5 = 3$ . What is $x$ ?

A
$x = 4$
B
$x = - 2$
C
$x = 14$
D
$x = 8$

Check your answer

Answer:

Before you read

Answer first — see what you already know.

A flow rate satisfies $\frac{12}{x} = 4$ . What is $x$ ?

A
$x = 48$
B
$x = 3$
C
$x = 8$
D
$x = 16$

Check your answer

Answer:

Theory

Radical Equations

A radical equation has a variable under a radical: $\sqrt{x + 3} = 5$ .

Steps to solve:
1. Isolate the radical on one side
2. Square both sides to eliminate the radical
3. Solve the resulting equation
4. CHECK your answer(s) — squaring can introduce extraneous solutions!

$\sqrt{2x+3} = 5$
$2x+3 = 25$ square both sides
$x = 11$ solve

Solving a radical equation

Example 1

Solve $\sqrt{x + 7} = 4$

Square both sides: $x + 7 = 16$

$x = 9$

Check: $\sqrt{9 + 7} = \sqrt{16} = 4$ ✓

x = 9

Example 2

Solve $\sqrt{2x - 1} = x - 2$

Square: $2x - 1 = (x-2)^2 = x^2 - 4x + 4$

$0 = x^2 - 6x + 5 = (x-1)(x-5)$

$x = 1$ or $x = 5$

Check $x = 1$ : $\sqrt{1} = 1$ but $x - 2 = -1$ . $1 \neq -1$ ✗ (extraneous!)

Check $x = 5$ : $\sqrt{9} = 3$ and $5 - 2 = 3$ ✓

x = 5

only

Theory

Rational Equations

A rational equation has a variable in a denominator: $\frac{3}{x} + 1 = \frac{5}{x}$ .

Steps:
1. Find the LCD (least common denominator)
2. Multiply every term by the LCD to clear fractions
3. Solve the resulting equation
4. CHECK — solutions that make a denominator $= 0$ are extraneous!

Example 1

Solve $\frac{3}{x} + 1 = \frac{5}{x}$

LCD is $x$ . Multiply all terms by $x$ :

$3 + x = 5$

$x = 2$

Check: $\frac{3}{2} + 1 = \frac{5}{2}$ $\to$ $\frac{5}{2} = \frac{5}{2}$ ✓

x = 2

Example 2

Solve $\frac{x}{x-2} = \frac{4}{x-2} + 1$

Multiply by $(x-2)$ : $x = 4 + (x-2)$

$x = x + 2$

$0 = 2$ — contradiction!

No solution.

No solution

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