Maths Game! (1 Viewer)

[bored.of.u]

Hi a new maths game ^^

Rules: Answer a question and then ask a question, anywhere from 2unit prelim to 3unit. Have fun guys.

I'll post the first question,

If, x + y=1 and x^3 + y^3=19 find the value of x^2 + y^2

 

[gurmies]

Strange question, never seen anything like it before. Nevertheless:

x + y = 1

x^3 + y^3 = 19

(x+y)^2 = x^2 + 2xy + y^2 = 1

x^2 + y^2 = 1 - 2xy

(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 = 1

3x^2y + 3xy^2 + 19 = 1

3x^2y + 3xy^2 = - 18

x^2y + y^2x = -6

xy(x + y) = -6

xy = -6 [x + y = 1]

So,

x^2 + y^2 = 1 - 2(-6)

= 13

Alternatively:

(x+y)^2 = x^2 + 2xy + y^2 = 1

x^2 + y^2 = 1 - 2xy [1]

x^3 + y^3 = (x+y)(x^2 - xy + y^2) = 19

x^2 - xy + y^2 = 19

x^2 + y^2 = 19 + xy [2]

Equating [1] & [2]

1 - 2xy = 19 + xy

3xy = -18

xy = -6

Subbing into [2], 19 - 6 = 13



New question: Find loci of |z + 1/z| = 2

 

[invad3r]

hi a new maths game ^^

rules: Answer a question and then ask a question, anywhere from 2unit prelim to 4unit. Have fun guys.

I'll post the first question,

if, x + y=1 and x^3 + y^3=19 find the value of x^2 + y^2

10

 

[bored.of.u]

10

Sory the answer is 13 gurmies your correct! =D

 

[azureus88]

hey gurmies, can you give us a clue for your question.

 

[vds700]

hey gurmies, can you give us a clue for your question.

its a 4 unit question

let z = x + yi, then simplify

 

[gurmies]

azureus88 it's a pretty interesting question, I suggest you do what vds700 said, but even then it's not easy to interpret the answer =)

 

[shady145]

hehe when i first saw the q i tried that but then my final answer was a straight line, and i wasn't confident i did it right. heres my answer anyway
y=-x+1.5

 

[gurmies]

Shady, I have a different answer to this question. Why don't you post up your solution and we'll try and find the flaw in it. If no flaws, i'll post up mine and you can find the flaws in it

 

[azureus88]

hehe when i first saw the q i tried that but then my final answer was a straight line, and i wasn't confident i did it right. heres my answer anyway
y=-x+1.5

when i did that, i ended up with powers of 6s and 4s and 2s

 

[gurmies]

Well here's my solution:

Spoiler

|z + 1/z| = 2

|z² + 1| = 2|z|

Now, z² = x² - y² + 2xyi

z² + 1 = (x² - y² + 1) + 2xyi

√[(x² - y² + 1)² + 4x²y²] = 2√(x² + y²)

(x² - y² + 1)² + 4x²y² = 4(x² + y²)

Spoiler

x⁴ + y⁴ + 1 - 2x²y² + 2x² - 2y² + 4x²y² = 4x² + 4y²
x⁴ + y⁴ + 2x²y² - 2x² - 2y² + 1 - 4y² = 0
(x² + y² - 1)² - 4y² = 0

(x² + y² -1 + 2y)(x² + y² - 1 - 2y) = 0

Therefore, the locus is the two circles:

x² + (y + 1)² = 2 and x² + (y - 1)² = 2

Those who wish to try it themselves please post up solution so we can compare =)

 

[Trebla]

If there are going to be questions ranging from 2 unit to Ext2, this thread should belong to the Ext2 forum...

 

[gurmies]

^ agreed

 

[bored.of.u]

^ agreed

haha ok i'll limit it to 3u then =D gurmies u wanna post up a new question?

 

[gurmies]

Alright, let's start things off simple then

Find in terms of t, where t = tan (x/2)

cos (x) + sin² (x/2)

 

[random-1005]

Well here's my solution:

Spoiler

|z + 1/z| = 2
|z² + 1| = 2|z|

Now, z² = x² - y² + 2xyi

z² + 1 = (x² - y² + 1) + 2xyi

√[(x² - y² + 1)² + 4x²y²] = 2√(x² + y²)

(x² - y² + 1)² + 4x²y² = 4(x² + y²)

x⁴ + y⁴ + 1 - 2x²y² + 2x² - 2y² + 4x²y² = 4x² + 4y²
x⁴ + y⁴ + 2x²y² - 2x² - 2y² + 1 - 4y² = 0
(x² + y² - 1)² - 4y² = 0

(x² + y² -1 + 2y)(x² + y² - 1 - 2y) = 0

Therefore, the locus is the two circles:

x² + (y + 1)² = 2 and x² + (y - 1)² = 2

Those who wish to try it themselves please post up solution so we can compare =)

i dont think this solution is correct lol, i havent done it for myself but looking through yours i dnt remember there being a property of
|z² + 1| = 2|z|
unless it was some sort of a mistake, trying to put |z + 1/z| over a common denominator of z
From line two to line three is wrong, u cannot write

z² + 1 = x² - y² + 1+ 2xyi (the x² - y² + 1 should not be put in brackets ), because in the third line u write that hence:
√[(x² - y² + 1)² + 4x²y²] = 2√(x² + y²)

which it does not, i can see that you have squared both the terms inside the brackets and then taken the sqrt of them, thinking that they cancel each other out, for this approach to work(ie be correct), the whole ( x² - y² + 1+ 2xyi) should have been squared, you cannot break them up into seperate parts and square them individually.

use a simple example
does (3+2+5)² +(4)² = (3+2+5+4)²
LHS=10²+16=116
RHS=14²= 196

we can see that this is not true

Additionally, when 2xyi is squared u get 4x²y²i² (i²=-1),
therefore = -4x²y² ( u have +4x²y²) be careful of that.

 

[random-1005]

Well here's my solution:

Spoiler

|z + 1/z| = 2

|z² + 1| = 2|z|

Now, z² = x² - y² + 2xyi

z² + 1 = (x² - y² + 1) + 2xyi

√[(x² - y² + 1)² + 4x²y²] = 2√(x² + y²)

(x² - y² + 1)² + 4x²y² = 4(x² + y²)

x⁴ + y⁴ + 1 - 2x²y² + 2x² - 2y² + 4x²y² = 4x² + 4y²
x⁴ + y⁴ + 2x²y² - 2x² - 2y² + 1 - 4y² = 0
(x² + y² - 1)² - 4y² = 0

(x² + y² -1 + 2y)(x² + y² - 1 - 2y) = 0

Therefore, the locus is the two circles:

x² + (y + 1)² = 2 and x² + (y - 1)² = 2

Those who wish to try it themselves please post up solution so we can compare =)

to be dead honest with you i have forgotten how to do these questions, however i am very sure it is not a straight line (as the modulus is the distance from the origin the locus must be a circle)

 

[gurmies]

Sorry but you have made a few mistakes yourself in recognising what it is that I have actually done, which is partly my fault to.

The locus to find out is:

|z + 1/z| = 2

|(z^2 +1)/z| = 2

|z^2 + 1| / |z| = 2

|z^2 + 1| = 2|z|

Using simple modulus or absolute value properties.

The next lines proceed to actually find the modulus of the equation I have above, where z = x + iy.

When you take the modulus of say, x + iy, you get √(x^2 + y^2) where your "i" theory is unnecessary, as we are finding a length...

 

[random-1005]

Well here's my solution:

Spoiler

|z + 1/z| = 2

|z² + 1| = 2|z|

Now, z² = x² - y² + 2xyi

z² + 1 = (x² - y² + 1) + 2xyi

√[(x² - y² + 1)² + 4x²y²] = 2√(x² + y²)

(x² - y² + 1)² + 4x²y² = 4(x² + y²)

x⁴ + y⁴ + 1 - 2x²y² + 2x² - 2y² + 4x²y² = 4x² + 4y²
x⁴ + y⁴ + 2x²y² - 2x² - 2y² + 1 - 4y² = 0
(x² + y² - 1)² - 4y² = 0

(x² + y² -1 + 2y)(x² + y² - 1 - 2y) = 0

Therefore, the locus is the two circles:

x² + (y + 1)² = 2 and x² + (y - 1)² = 2

Those who wish to try it themselves please post up solution so we can compare =)

LOL looking at it again i found another problem, even though i cant really follow how you go from
(x² + y² -1 + 2y)(x² + y² - 1 - 2y) = 0
to the two eqns of the circles
x² + (y + 1)² = 2 and x² + (y - 1)² = 2

remember that the form of a circle is (x-a)²+(y-b)²=r² with centre(a,b) and radius r
the radius in this case will be 2, as the modulus we want is 2

therefore IF your solns were correct, which im very sure they are not, the equations would be
(x-a)²+(y-b)²=4 (r²=4)

 

[gurmies]

I believe my solutions are correct, until somebody takes a logical and coherent stab at them. What you have said has made no sense...