Math Problem Statement

Solving radical equation

Solution

To solve a radical equation, follow these general steps:

  1. Isolate the radical term: Make sure that one side of the equation consists of a radical term (like a square root, cube root, etc.), and the other side contains only constants or variables.

  2. Eliminate the radical: Raise both sides of the equation to the power that matches the index of the radical. For example, if you have a square root, square both sides; if it's a cube root, cube both sides.

  3. Solve the resulting equation: Once the radical is eliminated, solve the resulting polynomial or algebraic equation.

  4. Check for extraneous solutions: Sometimes squaring both sides introduces solutions that don't actually satisfy the original equation. Substitute your solutions back into the original equation to verify their validity.

Example:

Solve the equation 3x+4=x2\sqrt{3x + 4} = x - 2.

Step 1: Isolate the radical term. 3x+4=x2\sqrt{3x + 4} = x - 2

Step 2: Square both sides to eliminate the square root. (3x+4)2=(x2)2(\sqrt{3x + 4})^2 = (x - 2)^2 3x+4=(x2)(x2)3x + 4 = (x - 2)(x - 2) Expand the right-hand side: 3x+4=x24x+43x + 4 = x^2 - 4x + 4

Step 3: Rearrange the equation. Move all terms to one side: 0=x24x3x+440 = x^2 - 4x - 3x + 4 - 4 0=x27x0 = x^2 - 7x

Factor the quadratic: x(x7)=0x(x - 7) = 0

Step 4: Solve for xx: x=0orx=7x = 0 \quad \text{or} \quad x = 7

Step 5: Check for extraneous solutions.

  • For x=0x = 0: Substitute into the original equation: 3(0)+4=024=2(False, discard x=0)\sqrt{3(0) + 4} = 0 - 2 \quad \Rightarrow \quad \sqrt{4} = -2 \quad (\text{False, discard } x = 0)
  • For x=7x = 7: Substitute into the original equation: 3(7)+4=7221+4=525=5(True, keep x=7)\sqrt{3(7) + 4} = 7 - 2 \quad \Rightarrow \quad \sqrt{21 + 4} = 5 \quad \Rightarrow \quad \sqrt{25} = 5 \quad (\text{True, keep } x = 7)

Thus, the solution is x=7x = 7.

Would you like any further details or have any questions about this solution?


Here are 5 related questions:

  1. How would you solve a radical equation with two radical terms?
  2. What happens if the equation involves a cube root instead of a square root?
  3. How do you solve an equation with both radicals and exponents?
  4. What are some common mistakes when solving radical equations?
  5. How can extraneous solutions arise, and how do you avoid them?

Tip: Always verify your solution in the original equation to rule out extraneous solutions!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Radical Equations
Quadratic Equations

Formulas

To eliminate a square root: (√a)^2 = a
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Fundamental Theorem of Algebra
Principle of Powers for Solving Radicals

Suitable Grade Level

Grades 9-12