Solving Radical Equations

Equations containing radicals

Solving Radical Equations

Strategy

Isolate the radical on one side
Raise both sides to the appropriate power
Solve the resulting equation
Check for extraneous solutions

Why Check?

Squaring both sides can introduce extraneous solutions.

Always substitute back into the original equation!

Multiple Radicals

If there are two radicals:

Isolate one radical
Square both sides
Isolate the remaining radical
Square again
Solve and check

Domain Restrictions

For $\sqrt{x}$ : must have $x \geq 0$

For $\sqrt[3]{x}$ : $x$ can be any real number

Example

Solve: $\sqrt{2x + 3} = 5$

Square both sides: $2x + 3 = 25$ $2x = 22$ $x = 11$

Check: $\sqrt{2(11) + 3} = \sqrt{25} = 5$ ✓

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve: $\sqrt{x + 5} = 4$

💡 Show Solution

Square both sides: $x + 5 = 16$

Solve for $x$ : $x = 11$

Check: $\sqrt{11 + 5} = \sqrt{16} = 4$ ✓

Answer: $x = 11$

2Problem 2medium

❓ Question:

Solve: $\sqrt{3x - 2} + 4 = 10$

💡 Show Solution

Step 1: Isolate the radical $\sqrt{3x - 2} = 6$

Step 2: Square both sides $3x - 2 = 36$

Step 3: Solve $3x = 38$ $x = \frac{38}{3}$

Check: $\sqrt{3(\frac{38}{3}) - 2} + 4 = \sqrt{38 - 2} + 4 = \sqrt{36} + 4 = 6 + 4 = 10$ ✓

Answer: $x = \frac{38}{3}$

3Problem 3hard

❓ Question:

Solve: $\sqrt{x + 7} = x - 5$

💡 Show Solution

Step 1: Square both sides $x + 7 = (x - 5)^2$ $x + 7 = x^2 - 10x + 25$

Step 2: Rearrange to standard form $0 = x^2 - 11x + 18$

Step 3: Factor $0 = (x - 9)(x - 2)$

Step 4: Solve $x = 9 \text{ or } x = 2$

Step 5: Check both solutions

For $x = 9$ : $\sqrt{9 + 7} = \sqrt{16} = 4$ and $9 - 5 = 4$ ✓

For $x = 2$ : $\sqrt{2 + 7} = \sqrt{9} = 3$ and $2 - 5 = -3$ ✗

Answer: $x = 9$ only (x = 2 is extraneous)

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