math t

[selina_934]

1.Express surd(59-(24surd6)) in the form (p surd 2 )+ (q surd 3) where p and q are integers

2.if (x+iy) square= i ,find all the real values of x and y.

3.using the laws of algebra of sets,show that,for any sets A and B,
(A-B)U(B-A)=(AUB)-(AnB)



ANS: 1.p=4,q=3
2.x=1/surd 2 ,y=1/surd 2
x= -1/surd 2,y= -1/surd 2

[Nicholasng925]

2.if (x+iy) square= i ,find all the real values of x and y.



Question 2: (x + iy)^2 = i
x^2 + 2xyi - y^2 = i

Compare the coefficient of i,
x^2 + y^2 = 0 -(2) ; 2xy = 1 ----> y = 1/2x -(1)

Substitute (1) into (2),

x^2 - 1/4x^2 = 0
4x^4 = 1
x^4 = 1/4
x^2 = 1/2
x = 1/surd 2 , -1/surd 2
y = 1/surd 2 , -1/surd 2 #

[selina_934]

Question 2: (x + iy)^2 = i
x^2 + 2xyi - y^2 = i

Compare the coefficient of i,
x^2 + y^2 = 0 -(2) ; 2xy = 1 ----> y = 1/2x -(1)

Substitute (1) into (2),

x^2 - 1/4x^2 = 0
4x^4 = 1
x^4 = 1/4
x^2 = 1/2
x = 1/surd 2 , -1/surd 2
y = 1/surd 2 , -1/surd 2 #



can straight away make x^4 bcome x^2???oo..i dunno that lo..that's y onli cannot get the ans....THANX A LOT..

[KenT13]

No.3
=(A^B')U(B^A')
=[(A^B')UB]^[(A^B')UA']
=[(AUB)^(BUB')]^[(A'UA)^(B'UA')] [ distributive law]
=[AUB^E]^[E^(B^A)'] [De morgan's and complement law]
=(AUB)^(B^A)'
=AUB-(A^B) [shown]

^ stands for "intersect"
E stands for universal set

Wow, your teacher is fast. Both of my Tt and school teachers haven't reach this part T.T

[selina_934]

No.3
=(A^B')U(B^A')
=[(A^B')UB]^[(A^B')UA']
=[(AUB)^(BUB')]^[(A'UA)^(B'UA')] [ distributive law]
=[AUB^E]^[E^(B^A)'] [De morgan's and complement law]
=(AUB)^(B^A)'
=AUB-(A^B) [shown]

^ stands for "intersect"
E stands for universal set

Wow, your teacher is fast. Both of my Tt and school teachers haven't reach this part T.T



ur teachers still haven't teach tis part but u oredi know how to do....u are too smart oredi!!!!!:)

[maxho5385]

If z=x+yi and z^2=a+bi, prove that 2x^2=sqrt(a^2+b^2) +a

By solving the equationb z^4 + 6z^2 +25=0 for z^2 or otherwise, express each root of the equation in the form of x+yi.

need help...thx:)

[markwongsk]

If z=x+yi and z^2=a+bi, prove that 2x^2=sqrt(a^2+b^2) +a

By solving the equationb z^4 + 6z^2 +25=0 for z^2 or otherwise, express each root of the equation in the form of x+yi.

need help...thx:)

z^2 = (x+yi)^2 = x^2 - y^2 + 2xyi

Since we are only concerned with x^2, compare the real part with z^2's real part, we get x^2 - y^2 = a.

Now, let z' be the conjugate of z, that is, let z' = x-yi and z^2' be the conjugate of z^2. Observe that (z*z')^2 = z^2*z'^2 and get an equation. I think it should be obvious after that.

[blublue]

selina 934

I think your answer for Question 1 is wrong....perhaps typing error....it suppose 2 be p=4; q= -3 rite? Can u check it out??

The working will be:-
let surd(59-(24surd6))=p(surd2)+q(surd3)
square both side will left only
59-24(surd6)=2p^2+3q^2+2pq(surd6)
Equate both side u will hav 2 equation
2p^2+3q^2=59.............(1)
-24=2pq
q=-12/p.................(2)
Sub (2) into (1)
2p^2+3(-12/p)^2=59
2p^2+432/(p^2)=59
2p^4-59p^2+432=0
Factorise the equation
(2p^2-27)(p^2-16)=0
p^2=27/2 and p^2=16
p= +/- surd (27/2)...not accepted and p=+/-4
Sub p into q
q=-/+3
Check answer by substituting value of p and q
value of p=-4; q=3 is not accepted since the answer is (-)ve value and square root of (-)ve value is not exist. Therefore only p=4; q=-3 is the answer. Hope it does help u...=)

[melindaL]

hey guys, can you please help me with these ques? prove the following
1) (surd5-surd3)/(surd5+surd3) - surd7/(surd7-surd5) + 5surd5/(4surd3 -2surd7) =1/2

2) (surd3+1/surd3-1)^3 - (surd3-1/surd3+1)^3 = 30surd3

thanks!

for question 1 :

surd(59-24surd6)^2 = (surdx -surdy)^2
59-24surd6 = x- 2surd (xy) + y

compare the surd part:
-24surd6 = -2surd(xy)
-12surd6 = -surd(xy)
(12surd6)^2 = (surd(xy))^2
144(6) = xy
xy = 864
y= 864/x ----(1)


compare the real part:
x+y =59
x+ (864/x) = 59
x^2-59x+864=0
(use the formula to find x)
x= 32 or x= 27
substitute x into equation (1)
y= 27 or y= 32
hence,
surd(59-24surd6) = +/- surd 32 - surd 27
= +/- surd( (16)(2)) - surd ((3)(9))
= 4surd2-3surd3 or -(4surd2-3surd3)
p=4, q=3

[markwongsk]

hey guys, can you please help me with these ques? prove the following
1) (surd5-surd3)/(surd5+surd3) - surd7/(surd7-surd5) + 5surd5/(4surd3 -2surd7) =1/2

2) (surd3+1/surd3-1)^3 - (surd3-1/surd3+1)^3 = 30surd3

thanks!

for question 1 :

surd(59-24surd6)^2 = (surdx -surdy)^2
59-24surd6 = x- 2surd (xy) + y

compare the surd part:
-24surd6 = -2surd(xy)
-12surd6 = -surd(xy)
(12surd6)^2 = (surd(xy))^2
144(6) = xy
xy = 864
y= 864/x ----(1)


compare the real part:
x+y =59
x+ (864/x) = 59
x^2-59x+864=0
(use the formula to find x)
x= 32 or x= 27
substitute x into equation (1)
y= 27 or y= 32
hence,
surd(59-24surd6) = +/- surd 32 - surd 27
= +/- surd( (16)(2)) - surd ((3)(9))
= 4surd2-3surd3 or -(4surd2-3surd3)
p=4, q=3

For question 2 use the identity (a-b)^3 = a^3 - b^3 -(a^2)b + a(b^2)

[PennyLing]

anyone please help me with bearing and 3D questions??