1. ## Inequality help?

I'm having trouble answering these questions, especially b) and c). Any help/answers would be greatly appreciated

maths.bmp

a and b.bmp

2. ## Re: Inequality help?

b) R.T.P. $\sqrt{2+\sqrt2} < \frac{6+\sqrt 2}{4}$

Squaring both sides,

$\mathrm{LHS} = 2 +\sqrt 2$

$\mathrm{RHS} = \frac{38 + 12\sqrt2}{16}$

Multiplying both sides by 16,

$\mathrm{LHS} = 32 + 16\sqrt 2$

$\mathrm{RHS} = 38 + 12\sqrt2$

Can you finish it from here?

c) seems to be missing some context... what are $a$ and $b$?

Also this looks like it would belong in the Mathematics subforum (i.e. the 2U one).

3. ## Re: Inequality help?

For part (b), literally follow the hint: after squaring both sides and simplifying, you should get
32 + 16√2 < 38 + 12√2,
which simplifies further to
√2 < 3/2,
which is the same inequality in part (a).

I can't help you with part (c), since I don't know what a and b are.

4. ## Re: Inequality help?

Originally Posted by fan96
b) R.T.P. $\sqrt{2+\sqrt2} < \frac{6+\sqrt 2}{4}$

Squaring both sides,

$\mathrm{LHS} = 2 +\sqrt 2$

$\mathrm{RHS} = \frac{38 + 12\sqrt2}{16}$

Multiplying both sides by 16,

$\mathrm{LHS} = 32 + 16\sqrt 2$

$\mathrm{RHS} = 38 + 12\sqrt2$

Can you finish it from here?

c) seems to be missing some context... what are $a$ and $b$?

Also this looks like it would belong in the Mathematics subforum (i.e. the 2U one).
Yes sorry this is in the wrong forum! But I'm a bit confused at how 6+√2/4 squared does not give 36+2/16? I've attached a picture above now showing the values for a and b. Sorry, I probably seem like a maths noob

5. ## Re: Inequality help?

Originally Posted by sida1049
For part (b), literally follow the hint: after squaring both sides and simplifying, you should get
32 + 16√2 < 38 + 12√2,
which simplifies further to
√2 < 3/2,
which is the same inequality in part (a).

I can't help you with part (c), since I don't know what a and b are.
Thanks for your help!! I've now attached values for a and b

6. ## Re: Inequality help?

Originally Posted by atamiabwv
how 6+√2/4 squared does not give 36+2/16?
$(a +b)^2 = a^2+2ab+b^2$

With $a = 6$, $b = \sqrt 2$:

$(6 + \sqrt 2)^2 = 6^2 + 6\sqrt2 + 6\sqrt 2 + 2^2 = 38 + 12\sqrt 2$

7. ## Re: Inequality help?

Originally Posted by fan96
$(a +b)^2 = a^2+2ab+b^2$

With $a = 6$, $b = \sqrt 2$:

$(6 + \sqrt 2)^2 = 6^2 + 6\sqrt2 + 6\sqrt 2 + 2^2 = 38 + 12\sqrt 2$
That makes more sense now, thank you heaps

8. ## Re: Inequality help?

Originally Posted by fan96
b) R.T.P. $\sqrt{2+\sqrt2} < \frac{6+\sqrt 2}{4}$

Squaring both sides,

$\mathrm{LHS} = 2 +\sqrt 2$

$\mathrm{RHS} = \frac{38 + 12\sqrt2}{16}$

Multiplying both sides by 16,

$\mathrm{LHS} = 32 + 16\sqrt 2$

$\mathrm{RHS} = 38 + 12\sqrt2$

Can you finish it from here?

c) seems to be missing some context... what are $a$ and $b$?

Also this looks like it would belong in the Mathematics subforum (i.e. the 2U one).
Do you perhaps have any idea how to do part (c) now, after the attachment?

9. ## Re: Inequality help?

Originally Posted by atamiabwv
Do you perhaps have any idea how to do part (c) now, after the attachment?
Given $\sqrt{2+\sqrt2}<\frac{6+\sqrt 2}{4}$

R.T.P. $4\sqrt{2-\sqrt2} < 8\sqrt{2-\sqrt{2+\sqrt 2}}$

Square both sides

$LHS = 16(2-\sqrt 2)$

$RHS = 64(2-\sqrt {2+\sqrt 2})$

Divide both sides by 64:

$LHS = \frac{2-\sqrt 2}{4}$

$RHS = 2-\sqrt {2+\sqrt 2}$

This is starting to look like the result of part b). Can you take it from here?

10. ## Re: Inequality help?

Originally Posted by fan96
Given $\sqrt{2+\sqrt2}<\frac{6+\sqrt 2}{4}$

R.T.P. $4\sqrt{2-\sqrt2} < 8\sqrt{2-\sqrt{2+\sqrt 2}}$

Square both sides

$LHS = 16(2-\sqrt 2)$

$RHS = 64(2-\sqrt {2+\sqrt 2})$

Divide both sides by 64:

$LHS = \frac{2-\sqrt 2}{4}$

$RHS = 2-\sqrt {2+\sqrt 2}$

This is starting to look like the result of part b). Can you take it from here?
That's really helpful, thank you very much! I've taken it a bit further for the LHS, but I'm not sure what do to for the RHS. I've attached my working: IMG_6433.bmp

11. ## Re: Inequality help?

Originally Posted by atamiabwv
That's really helpful, thank you very much! I've taken it a bit further for the LHS, but I'm not sure what do to for the RHS. I've attached my working: IMG_6433.bmp
Don't square it - you want to get it to the result in part b).

You'll be able to see the answer by doing just one subtraction.

12. ## Re: Inequality help?

Originally Posted by fan96
Don't square it - you want to get it to the result in part b).

You'll be able to see the answer by doing just one subtraction.
Hmm I’ve tried subtracting -2 from both sides, other than that I don’t know what else I could really subtract. I’m still not getting the answer, and my classmates are confused too. Sorry for all the hassle

13. ## Re: Inequality help?

$LHS = \frac{2-\sqrt 2}{4}$

$RHS = 2-\sqrt {2+\sqrt 2}$

Subtract 2:

$LHS = \frac{-6-\sqrt 2}{4}$

$RHS = -\sqrt {2+\sqrt 2}$

Multiply by -1 (which flips the inequality sign)

$LHS = \frac{6+\sqrt 2}{4}$

$RHS = \sqrt {2+\sqrt 2}$

Which is the result in b)

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