Radical inequation

kiriku123

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Mar 24, 2019
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Could someone help me about this problem, you should find values for x:

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Jomo

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Bring the sqrt(x+3) to the right hand side and then square both sides. Show us your work and then we will proceed from there.
 

kiriku123

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Mar 24, 2019
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yeah nvm i solved it
 

Dr.Peterson

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You have to be very careful if you square an inequality. It's probably better to make it an equation and find whether there are any points where they are equal, and then consider intervals between those points. That doesn't seem easy, though.

On the other hand, if you just look at the inequality after subtracting \(\displaystyle \sqrt{x+3}\) and think long enough about signs, the solution may become clear.

Or you could just graph the LHS to see what it looks like.
 

Jomo

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From the domain we get that -3 < x < 9. A quick inspection show that 0 < x < 9 are solution. The question is which solutions are there between -3 and 0.
 
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Dr.Peterson

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Looking at the inequality, \(\displaystyle \sqrt{x+3} + \sqrt[4]{9-x} > \sqrt{3}\), we can consider two cases. If x > 0, the first term alone is greater than the RHS, and the second is positive (unless x = 9), so the inequality is true. On the other hand, if x < 0, the second term alone is greater than the RHS, and the first is positive (unless x = -3), so again it is true.
 
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