7.5 – Solving Radical Equations

I. Solving Radical Equations Radical Equation – an equation that has a variable in the radicand or a variable with a rational exponent.Radical Equation – an equation that has a variable in the radicand or a variable with a rational exponent. √x + 3 = x 1/2 = 17

Steps to Solving Rational Equaitons:Steps to Solving Rational Equaitons: –Step 1: isolate the radical or radical expression on one side –Step 2: raise both sides of the equation to the same power –Step 3: if you have a rational exponent, you can multiply each side by the reciprocal of the exponent to achieve an exponent of 1. –Step 4: check for extraneous solutions. They can arise when you raise both sides of the equation to a specified power.They can arise when you raise both sides of the equation to a specified power.

Example 1: solve the following:Example 1: solve the following: A) 2 + √(3x – 2) = 6A) 2 + √(3x – 2) = 6 B) √(5x + 1) – 6 = 0B) √(5x + 1) – 6 = 0 C) 3√x + 3 = 15C) 3√x + 3 = 15 D) √(3x + 4) = 4D) √(3x + 4) = 4

Example 2: solve the followingExample 2: solve the following A)2(x – 2) 2/3 = 50 B)2(x + 3) 3/2 = 54 C)3 + (4 – x) 3/2 = 11

Example 3: check for extraneous solutionsExample 3: check for extraneous solutions A)√(x – 3) + 5 = x B)√(5x – 1) + 3 = x C)√(x + 7) + 5 = x

If your equation contains two radical expressions or rational exponents, isolate one of the radicals or one of the terms, and solve like before.If your equation contains two radical expressions or rational exponents, isolate one of the radicals or one of the terms, and solve like before. Check for extraneous solutionsCheck for extraneous solutions

Example 4: solve the following:Example 4: solve the following: A)(2x + 1).5 – (3x + 4).25 = 0 A)√(3x + 2) - √(2x + 7) = 0 B)√(x + 10) + √(3 – x) = 5