Cambridge 3u Maths problem. (1 Viewer)

ieatbarbies · Mar 8, 2009

Would anyone know how to solve from the Cambridge Yr11 3u book ex 1B q15 & 16?

Ughh wish I had access to the worked solutions

monesh · Mar 8, 2009

Why not just write the questions out here, then others could help...

jet · Mar 8, 2009

$2x = a + b + c \\ \text{LHS} = (x - a)^2 + (x - b)^2 + (x - c)^2 + x^2 \\ = 4x^2 -2x(a + b + c) + a^2 + b^2 + c^2 \text{ upon expansion} \\ = 4x^2 - 2x(2x) + a^2 + b^2 + c^2 \\ = 4x^2 -4x^2 + a^2 + b^2 + c^2 \\ = a^2 + b^2 + c^2 = \text{RHS} \\ \\ LHS = (a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd) \text{ by the question}\\ a^2 + 2b^2 + 2c^2 + d^2 + 2(ab + bc + cd) = 4(ab + bc + cd) \\ \therefore a^2 + 2b^2 + 2c^2 + d^2 = 2(ab + bc + cd) \\ (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 -2cd + d^2) = 0 \\ (a - b)^2 + (b - c)^2 + (c - d)^2 = 0 \\ \text{Now, by thinking about it, you will realise that all perfect squares must be positive (excluding complex numbers)} \\ \text{And so, the only way this inequality can occur is if} (a - b^2) = 0, (b - c)^2 = 0, (c - d)^2 = 0 \\ \therefore a = b, b = c, c = d \\ \therefore a = b = c = d$

Hope those were the right questions...

ieatbarbies · Mar 8, 2009

monesh said:
Why not just write the questions out here, then others could help...

Sry, asked this pretty late into the night half asleep, tired out from maths revisions. Should have, shouldn't I? Will do next ttime, thanks.

jetblack2007 said:
$2x = a + b + c \\ \text{LHS} = (x - a)^2 + (x - b)^2 + (x - c)^2 + x^2 \\ = 4x^2 -2x(a + b + c) + a^2 + b^2 + c^2 \text{ upon expansion} \\ = 4x^2 - 2x(2x) + a^2 + b^2 + c^2 \\ = 4x^2 -4x^2 + a^2 + b^2 + c^2 \\ = a^2 + b^2 + c^2 = \text{RHS} \\ \\ LHS = (a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd) \text{ by the question}\\ a^2 + 2b^2 + 2c^2 + d^2 + 2(ab + bc + cd) = 4(ab + bc + cd) \\ \therefore a^2 + 2b^2 + 2c^2 + d^2 = 2(ab + bc + cd) \\ (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 -2cd + d^2) = 0 \\ (a - b)^2 + (b - c)^2 + (c - d)^2 = 0 \\ \text{Now, by thinking about it, you will realise that all perfect squares must be positive (excluding complex numbers)} \\ \text{And so, the only way this inequality can occur is if} (a - b^2) = 0, (b - c)^2 = 0, (c - d)^2 = 0 \\ \therefore a = b, b = c, c = d \\ \therefore a = b = c = d$

Hope those were the right questions...

Thanks so much!! Repped

joben · Mar 8, 2009

jetblack2007 said:
$2x = a + b + c \\ \text{LHS} = (x - a)^2 + (x - b)^2 + (x - c)^2 + x^2 \\ = 4x^2 -2x(a + b + c) + a^2 + b^2 + c^2 \text{ upon expansion} \\ = 4x^2 - 2x(2x) + a^2 + b^2 + c^2 \\ = 4x^2 -4x^2 + a^2 + b^2 + c^2 \\ = a^2 + b^2 + c^2 = \text{RHS} \\ \\ LHS = (a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd) \text{ by the question}\\ a^2 + 2b^2 + 2c^2 + d^2 + 2(ab + bc + cd) = 4(ab + bc + cd) \\ \therefore a^2 + 2b^2 + 2c^2 + d^2 = 2(ab + bc + cd) \\ (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 -2cd + d^2) = 0 \\ (a - b)^2 + (b - c)^2 + (c - d)^2 = 0 \\ \text{Now, by thinking about it, you will realise that all perfect squares must be positive (excluding complex numbers)} \\ \text{And so, the only way this inequality can occur is if} (a - b^2) = 0, (b - c)^2 = 0, (c - d)^2 = 0 \\ \therefore a = b, b = c, c = d \\ \therefore a = b = c = d$

Hope those were the right questions...

Hey how did u get that font? is it a screenshot from a pdf? if so, can i have the worked solutions pdf please?

micuzzo · Mar 8, 2009

^^latex equation editor

khorne · Mar 8, 2009

That "font" is called LaTeX, and is a a typeset for doing equations and other non standard letters/numerals.

You can try it out, the button to launch the browser version is right above the quick reply bok in the mathematics sections.

jet · Mar 8, 2009

The 'font' is latex - the link to the editor is below. I know the code, and so it is very useful, though there is a graphical editor, which helps you to learn.
The only bad thing about the editor is that there is a character limit, which gets very annoying.

Cambridge 3u Maths problem. (1 Viewer)

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