Mega Maths Question to test your knowledge (1 Viewer)

[Real Madrid]

Solve:
(x^2-x)^2+(x^2-x)-2=0

Show working

 

[-tal-]

I got x = 1, x = -2 or x = (1 + sqrt5)/2 or x = (1 - sqrt5)/2

 

[shaon0]

Solve:
(x^2-x)^2+(x^2-x)-2=0

Show working

(x^2-x)((x^2-x)-2)=0
(x^2-x-1)(x^2-x+2)=0

Thus, x=0.5(1+sqrt(5)), 0.5(1-sqrt(5)), 0.5(1-i.sqrt(7)), 0.5(1+i.sqrt(7))
Sorry for my solution before.

 

[tommykins]

Solve:
(x^2-x)^2+(x^2-x)-2=0

let m = (x²-x)

m² + m - 2 = 0
(m+2)(m-1) = 0
(x²-x+2)(x² - x -1) = 0

(1) (x²-x+2) = 0
x = 1 +-sqrt[1²-4.2.1]/2
No solution as discriminant < 0.

(2) (x² - x -1) = 0
x = 1+-sqrt[1 + 4.1]/2
= (1+-sqrt5)/2

 

[Real Madrid]

Congrats you get 3/3

 

[bored of sc]

Solve:
(x^2-x)^2+(x^2-x)-2=0

Show working

As tommykins showed, reduce the equation to a quadratic by making a substitution. That it the safest way to do it I reckon.

 

[tommykins]

Congrats you get 3/3

YAYY

 

[lyounamu]

Solve:
(x^2-x)^2+(x^2-x)-2=0

Show working

Are you serious?

This is like...

 

[bawd]

Are you serious?

This is like...

Easy?

 

[Real Madrid]

Its a challenge in itself.

 

[lyounamu]

Its a challenge in itself.

Well, to express my own opinion on this, this is not a challenging question at all once you understand that you can create a separate entity to which x^2-x can be equal to.

 

[Real Madrid]

I had a brain freeze.

I'm worn out.

3 essays today in the exams!!!

 

[tommykins]

Here is a question which I'm sure you'll all enjoy.

Find cos(5x).

Hence find the solutions to
32x^5 - 40x^3 + 10x - sqrt3 = 0

 

[shaon0]

(cos(x)+i.sin(x))^5
Let c=cosx and s=sinx
(cos(x)+i.sin(x))^5=(c+s.i)^5
= c^5 + 5c^4*s.i - 10c^3s^2 - 10c^2*s^3i + 5c.s^4 + s^5i
cos(5x) = c^5 - 10c^3*s^2 + 5c.s^4
since; s^2=1-c^2
cos(5x) = c^5 - 10c3(1-c^2) + 5c (1-c^2)^2
= 11c^5 - 10c^3 + 5c(1-2c^2+c^4)
= 16c^5 - 20c^3 + 5c
2*cos(5x)=32c^5 - 40c^3 +10c
Let c=x.
2*cos(5x)=32x^5 - 40x^3 +10x
Let 2*cos(5x)= sqrt(3)
Thus, sqrt(3) = 32x^5 - 40x^3 +10x
32x^5 - 40x^3 + 10x - sqrt3 = 0

LOL, i don't think my solutions correct.

 

[tommykins]

(cos(x)+i.sin(x))^5
Let c=cosx and s=sinx
(cos(x)+i.sin(x))^5

Keep goin

 

[bored of sc]

Here is a question which I'm sure you'll all enjoy.

Find cos(5x).

Hence find the solutions to
32x^5 - 40x^3 + 10x - sqrt3 = 0

cos(3x+2x)
= cos3xcos2x - sin3xsin2x
= [cos(2x+x) . (cos²x - sin²x)] - [sin(2x+x) . 2sinxcosx]
= [cos2xcosx - sin2xsinx . (cos²x - sin²x)] - [(sin2xcosx + cos2xsinx) . 2sinxcosx]
= {[(cos²x - sin²x) cosx - 2sinxcosx] . [cos²x - sin²x]} - {[2sinxcosx + (cos²x - sin²x) sinx] . 2sinxcosx}

Does some more expanding...

= cos⁵x - sin²x.cos³x - 2sinx.cos³x - 4sin²x.cos²x + 2sin³.cosx + 2sin²x.cosx

Gets more confused.

 

[tommykins]

cos(3x+2x)
= cos3xcos2x - sin3xsin2x
= [cos(2x+x) . (cos²x - sin²x)] - [sin(2x+x) . 2sinxcosx]
= [cos2xcosx - sin2xsinx . (cos²x - sin²x)] - [(sin2xcosx + cos2xsinx) . 2sinxcosx]
= {[(cos²x - sin²x) cosx - 2sinxcosx] . [cos²x - sin²x]} - {[2sinxcosx + (cos²x - sin²x) sinx] . 2sinxcosx}

Does some more expanding...

= cos⁵x - sin²x.cos³x - 2sinx.cos³x - 4sin²x.cos²x + 2sin³.cosx + 2sin²x.cosx

Gets more confused.

Haha i know, I was a little mean.

 

[lolokay]

Here is a question which I'm sure you'll all enjoy.

Find cos(5x).

Hence find the solutions to
32x^5 - 40x^3 + 10x - sqrt3 = 0

cos5x
= Re(cos[x] + isin[x])⁵
= c⁵ - 10c³s² + 5cs⁴
substituting 1-c² for s² and expanding + simplifying gives:
= 16c⁵ - 20c³ + 5c

32x^5 - 40x^3 + 10x - sqrt3 = 0
let x = cos[a]
2cos[5a] = sqrt3
cos[5a] = sqrt3 /2 = +-cos[pi/12 +- 2npi)
a = +-pi/60 +-2/5 npi

 

[tommykins]

cos5x
= Re(cos[x] + isin[x])⁵
= c⁵ - 10c³s² + 5cs⁴
substituting 1-c² for s² and expanding + simplifying gives:
= 16c⁵ - 20c³ + 5c

32x^5 - 40x^3 + 10x - sqrt3 = 0
let x = cos[a]
2cos[5a] = sqrt3
cos[5a] = sqrt3 /2 = +-cos[pi/12 +- 2npi)
a = +-pi/60 +-2/5 npi

Very well done. Are you currently self teaching 4unit?

 

[Real Madrid]

Solve the attachment