# Inequalities, need some help please.

1. Jan 16, 2013

### christian0710

1. The problem statement, all variables and given/known data

Hi, i have a question regarding inequalities:
I have to solve this inequality: 4x^4 +(√2y)^4 ≤64

My first first quess was to cancel out the square root (√2y)^2*(√2y)^2
4x^4 + 4y^2 ≤64

Then take the square root on both sides (here my question is: do you take one big square root that covers both terms on the left, or of each single term as i do next?)

2x^2 + 2y ≤ 8

Now my question is, what hapens if i want to solve it with respect to x?? Does the inequality sign change if i subtract y?

2. Jan 16, 2013

### Staff: Mentor

What you wrote in the 2nd term on the left is $\sqrt{2}y$. Did you mean $\sqrt{2y}$? If so, you should write it as √(2y), with parentheses.
To answer your question, you have to take the square root of the whole side, not of the individual terms. What you are saying is that $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$, which is NOT true.

3. Jan 16, 2013

### christian0710

I see and yes you are right about meaning √(2y)^2 (my mistake)

The actual problem is one of a double integral - find the area D bounded by 4x^4 +(√(2y))^4 ≤64 , y≥0 I thought it was enough to just post the question regarding the inequalities, but I i see that i should probably include the whole context (sory)

Still, I think all i have to do is to isolate x right? because y is already defined as y≥ 0

So i move x to the right (subtract) and move 8 to the left 4y^2 - 8 ≤ 4x^4 but still i'm confused about if I must to isolate x (and how?) I've never worked with inequalities before now (which is a bit late)

4. Jan 16, 2013

### Staff: Mentor

It might be easier to solve for y than to solve for x. Adding and subtracting quantities to both sides of an inequality is straightforward, and doesn't change the direction of the inequality. It's only when you multiply or divide by a negative number that the direction changes.

You have a mistake - your inequality has "-8" that shouldn't be there. To isolate the y term, the steps look like this:
4x4 + 4y2 ≤ 64
x4 + y2 ≤ 16
y2 ≤ 16 - x4
y ≤ $\sqrt{16 - x^4}$

The above is simpler than usual because you are given that y ≥ 0.

In contrast, if you had this inequality: x2 ≤ 4, you have to take into account that x can be negative, so the solution is -2 ≤ x ≤ 2.

5. Jan 16, 2013

### Ray Vickson

You cannot do what you attempted above: the square root of 4x^4 + 4y^2 is not 2x^2 + 2y; in fact, √(a+b) ≠ √a + √b. For example, √8 = √(4+4) = 2.828427124, but √4 + √4 = 2+2 = 4.

6. Jan 16, 2013

### christian0710

Now i understand, so it's only division or multiplication by negative number that changes the signs, so in the last case x2 ≤ 4, y you just add an inequality sign because you know it's a ± solution since you isolated ONE variable that was raised to the second power :)

This may sound dumb, but what if you subtract y2 in y2 ≤ 16 - x4 then you get 0 ≤ 16 - x4 (so you put in a zero?)

Thank you so much!!

7. Jan 16, 2013

### Staff: Mentor

It's a bit more complicated than that, as it involves solving quadratic inequalities.

The example I gave was x2 ≤ 4, with solution -2 ≤ x ≤ 2.
If it had been x2 ≥ 4, the solution set is x ≤ -2 U x ≥ 2.
If you subtract a term from a side that has only that term, yes, you get 0 on that side, BUT you have to subtract if from the other side as well, so you would have 0 ≤ 16 - x4 - y2.

8. Jan 16, 2013

### christian0710

I see so perhaps quadratic inequalities is the next thing i'll learn :D
Thank's again, I appreciate your help.

9. Jan 16, 2013

### christian0710

Ah yes one very last question: according to the informating regarding the area D bound by 4x^4 +(√(2y))^4 ≤64 , y≥0 then this must be true no?
√(16-x^4)≥ y ≥0

So i know what y lies between, but would i also have to solve the inequality for x (this integral problem is bugging me a bit) in order to draw the region? Ahh wait the square root can't be negative so -2 ≤x^4 ≤ 2
so x must lie between -2^(1/4) ≤x ≤ 2^(1/4) am i right? :)

10. Jan 16, 2013

### Joffan

No, you're not right, although there are elements of correctness there. 16=2^4.