How to Solve Inequality with Two Square Roots Terms

[@ibrah1324]

It's always the Indian guy

[@guadalajara4848]

THIS IS TOTALLY WRONG : when you arrive at 7/4 - 2x > V(x + 1), you cannot just square both parts of the inequality without checking the sign of ( 7/4 - 2x ) : if it is negative (that is : x > 7/8) while the square root of the second part is (always) positive, you cannot say anything about what the inequality becomes after squaring : is it > or < ? Indeed, when you square -6 < 1, you don't get 36 < 1 but 36 > 1 ! If you take this into account, before being able to square and proceed like you do later, you have to force x to be less or equal to 7/8 ; and when you consider the opposite case (x > 7/8), you immediately notice that the inequality becomes absurd (a negative number being greater than a positive one - the square root). As a conclusion, you must reduce the range of your solution to x being less or equal to 7/8 !

[@VivaanArora-lv8jg]

bro your right

[@uttkarsh7823]

indeed you're true

[@narayananarayanakumar1466]

❤❤❤

[@yourmathtutorvids]

Great video. Was trying to do some refreshers on inequalities and cleared up the concept. Thank you!

[@ananthpadmanaban650]

Thanks for this sir . Between 1.6959 and 3 if you take a number 2and check for inequality you get absurd result. Hence this part may not be part of the solution.

[@gloryrina8518]

Very nice sir

[@lovemorembasela6491]

Kindly solve this inequality for me √1-2x>|2x+1|. I have had a great deal of trouble handling it

[@touhami3472]

Domain of validity is WRONG let p number > 0. sqrt(A) > p + sqrt(B) A > p^2 + B + 2psqrt(B) So A-B-p^2 > 2psqrt(B) { A-B-p^2 >=0 , B>=0 ==> D of validity (A-B-p^2)^2 > 4p^2B IN D } Here, A=3-x , B=x+1 , p=1/2 D of validity : B>=0 ===> x>=-1 and A-B>=p^2 ====> x

[@jenniferaboufadle9527]

But should it have been a comma or union U ? [-1,0.304)U(1.6959,3) ?

[@MathematicsTutor]

Union is better. Thanks

[@JayNadiminti-416]

Right side of union is not a part of the solution...you can check with "x =2".

[@bh5169]

Thank you !

[@ahhhwae5403]

Hi is it the same way to solve this equation but we just want the solution to be integer only? Or we have to set the range to be integer too?

[@MathematicsTutor]

Domain and Range is in real numbers. Some similar questions may have integer solutions. The method is exactly same. Thanks

[@gayathirigayathiri6732]

Square root under c+5 + square root under c+10 > 2 sir how to solve this problem

[@SuperSatyasrinivas]

Good method

[@anggalol]

Wow, this is actually an IMO question, nice

[@mfaecs]

Ans won't include (1.6959,3] I think. As in Step 4 which is (7/4 -2x)>root(x+1) any value from set (1.6959,3] will assign (7/4 -2x) a negative value. So you are basically saying (a negative value)> root(x+1) which is not possible as root is always positive.

[@eduardoalvarez163]

Thats correct. The solution is exactly [-1,1-sqrt(31)/8) = [-1,0.30403). You must evaluate the new restriction before power to 2 the root square.

[@touhami3472]

Domain would be: x+1>=0 and 3-x-(x+1) >= (1/2)^2 So D= [ -1 , 7/8 ] Therefore solution is [-1, 1-sqrt(31)/8 ) .

[@mylord7848]

I was thinking the same thing, the second time he squared both sides, I was confused because we cannot just do that as one side is certainly positive and the other side can be both positive and negative!

[@satpalsingh5197]

Maha gussa

[@hiteshkumar5356]

gussa kis liye bhai ?