√(x+6)>√(x-1)+√(2 x-5)

Name: Problem 254, Chapter 7: Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry
Uploaded: 2022-02-28T06:51:08-08:00
Duration: 4 min 28 s
Channel: Kavita Kumari
Description: √(x+6)>√(x-1)+√(2 x-5)

Key Concepts

Squaring Both Sides

Eliminating radicals by squaring both sides of an equation or inequality is a common technique. However, when applying this method, one must do so with caution because the process can introduce extraneous solutions and potentially alter the structure of the inequality. This concept emphasizes the importance of verifying any obtained solutions in the original inequality.

Extraneous Solutions

Extraneous solutions arise from algebraic manipulations, such as squaring both sides, that are not valid for the original equation or inequality. Recognizing and eliminating these solutions is essential to ensure that the final answers adhere to the initial conditions and domain restrictions of the problem.

Domain Restrictions

In problems involving square root expressions, it's crucial to ensure that the radicands (the expressions inside the square roots) are non-negative. This concept establishes the valid range of the variable values before any manipulation begins, as violating these restrictions can lead to undefined expressions in the context of real numbers.

Radical Inequalities

Inequalities that include radical expressions require careful handling since the properties of square roots and the nature of inequalities necessitate special procedures. This concept involves understanding how the behavior of square roots affects the inequality and dictates a systematic approach to solve it.

Transcript

00:01 In the following question, the equation, the inequality that is given to us is root x plus 6 greater than root x minus 1 plus root 2x minus 5.

00:15 Now firstly we need to know a very basic rule in case of equal inequalities that if root of a function, let's say fx, is greater than another function, then.

00:33 Then g x, which is the other function, will be greater than 0 or there can also be another situation where gx can be greater than or equal to 0 as well.

00:45 So this is one simple rule that we need to remember.

00:49 Let's solve our question further.

00:51 Taking the first situation where gx will be greater than 0, so our gx here is root x minus 1 plus root 2x minus 5, will be greater than 0.

01:04 This is our first situation.

01:06 So when we try to solve this out, we come to know that there's no solution that exists to this inequality.

01:14 So we can ignore this and move on to the second situation.

01:17 That is why we wrote down the second situation, because there's a possibility that we do not get the situation, we do not get an answer for the first case.

01:25 So moving on to the second case, this is going to be the inequality that we will form and now here we can be clear we can clearly see that there is a solution that is possible and if i write down the solution it will be root minus 149 minus 5 upon 2 and x will be greater than this term and less than root 149 minus 5 by 2 so this is the first we can say first interval that we obtain.

02:04 Now there's one more thing that we need to keep in mind...

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