How to Solve Radical Equations - We use exponent and basic algebraic laws to solve such equations. Plus, we look out blunders that caused by levitation unidentified values to an even exponent. Assumed equation: 3 + sqrt(5x + 6) = 2 Applied opposite signs of the same quantity as they will minus by each other. Then, take 2(power) on both sides to eradicate sqrt. {sqrt(5x+6)} 2(power) = (9)2(power) | 5x + 6 = 81 Opposite signs will minus. 5x/5 = 75/5 We will cut 5/5 by each other and x become isolate. Ans: x = 15 Verification: So, what will you do to confirm that is your final answer is correct or not? You have to do it by executing the verification method. For this, you have to put the value of x i.e. 15 in your assumed question. How will it be? Let us understand! 3 + sqrt(75+6) = 12 | 3 + sqrt(81) = 12 | 3 + 9 = 12. It is the point where it is vital that you must have the same answer or value on both sides. So, 3 + 9= 12. The final answer will be: 12 = 12 The value of x = 12 and verification value is 12. It means you had found the value of assumed question correct.

• ### Basic Lesson

Guides students solving equations that involve an Radical Equations. Demonstrates answer checking. Replace the inequality symbol with an equal sign and solve the resulting equation.

• ### Intermediate Lesson

Demonstrates how to solve more difficult problems.

• ### Independent Practice 1

A really great activity for allowing students to understand the concepts of Radical Equations.

• ### Independent Practice 2

Students find a series of Radical Equations in assorted problems. The answers can be found below.

• ### Homework Worksheet

Students are provided with problems to achieve the concepts of Radical Equations.

• ### Skill Quiz

This tests the students ability to evaluate Radical Equations.