# Inequality help? (1 Viewer)

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#### fan96

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

Squaring both sides,

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

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

Multiplying both sides by 16,

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

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

Can you finish it from here?

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

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

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#### sida1049

##### Well-Known Member
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.

#### atamiabwv

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

Squaring both sides,

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

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

Multiplying both sides by 16,

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

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

Can you finish it from here?

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

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

#### atamiabwv

##### New Member
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

#### fan96

##### 617 pages
how 6+√2/4 squared does not give 36+2/16?
$\bg_white (a +b)^2 = a^2+2ab+b^2$

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

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

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#### atamiabwv

##### New Member
$\bg_white (a +b)^2 = a^2+2ab+b^2$

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

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

#### atamiabwv

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

Squaring both sides,

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

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

Multiplying both sides by 16,

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

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

Can you finish it from here?

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

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?

#### fan96

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

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

Square both sides

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

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

Divide both sides by 64:

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

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

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

#### atamiabwv

##### New Member
Given $\bg_white \sqrt{2+\sqrt2}<\frac{6+\sqrt 2}{4}$

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

Square both sides

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

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

Divide both sides by 64:

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

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

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: View attachment IMG_6433.bmp

#### fan96

##### 617 pages
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: View attachment 34519
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.

#### atamiabwv

##### New Member
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

#### fan96

##### 617 pages
$\bg_white LHS = \frac{2-\sqrt 2}{4}$

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

Subtract 2:

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

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

Multiply by -1 (which flips the inequality sign)

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

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

Which is the result in b)