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Solve This Mathematics Problem by Johnpaul01: 12:08pm On Jan 17, 2017
Help me out, its an assignment and am getting confused
find the positive square root of
35-12√6
Re: Solve This Mathematics Problem by laperson: 4:48pm On Jan 17, 2017
The positive square root is 3√3-2√2

I will send the solution to you ASAP

Regards
Re: Solve This Mathematics Problem by laperson: 5:03pm On Jan 17, 2017
I couldn't upload the solution, don't understand what is happening-I didn't exceed the 200Kb limit for files. I will try to type the solution for you but that might take some time.

Regards
Re: Solve This Mathematics Problem by laperson: 5:21pm On Jan 17, 2017
Solution:
Let the square root of 35-12√6 be represented by √a - √b

Therefore, the 35-12√6 = (√a - √b)square= a+b-2√ab

comparing the equation and quantities above, we can clearly observe that

a+b=35----------- (1)

2√ab=12√6
√ab=6√6
square bothsides
ab=216------------(2)

Solving equations (1) and (2) simultaneously,

let a=35-b---------(3)

put (3) into (2), we obtain

(35-b)(b)=216
exploring further
36b-b[/sup]2=216
rearranging in quadratic form will give
b[sup]
2-35b+216=0
solving by factorisation method
b[sup][/sup]-27b-8b+216=0
b(b-27)-8(b-27)=0
b=8 or 27

However, when b=8, a=35-b=35-8=27 (a=27 and b=cool case 1
when b=27, a=35-b=35-27=8 (a=8 and b=27) case 2

Let us now explore case 1: a=27 and b=8

Note that the square root= √a - √b =√27 - √8= 3√3 - 2√2...This is positive because 3√3 - 2√2>0

but if you consider case 2: a=8 and b=27
The square root becomes the reverse i.e 2√2 - 3√3...This is negative because 2√2 - 3√3<0

Therefore, the positive square root is 3√3 - 2√2 and it is obtained when a=27 and b=8 in case 1 above

Regards
Re: Solve This Mathematics Problem by Johnpaul01: 7:48pm On Jan 17, 2017
laperson:
Solution:
Let the square root of 35-12√6 be represented by √a - √b

Therefore, the 35-12√6 = (√a - √b)square= a+b-2√ab

comparing the equation and quantities above, we can clearly observe that

a+b=35----------- (1)

2√ab=12√6
√ab=6√6
square bothsides
ab=216------------(2)

Solving equations (1) and (2) simultaneously,

let a=35-b---------(3)

put (3) into (2), we obtain

(35-b)(b)=216
exploring further
36b-b[/sup]2=216
rearranging in quadratic form will give
b[sup]
2-35b+216=0
solving by factorisation method
b[sup][/sup]-27b-8b+216=0
b(b-27)-8(b-27)=0
b=8 or 27

However, when b=8, a=35-b=35-8=27 (a=27 and b=cool case 1
when b=27, a=35-b=35-27=8 (a=8 and b=27) case 2

Let us now explore case 1: a=27 and b=8

Note that the square root= √a - √b =√27 - √8= 3√3 - 2√2...This is positive because 3√3 - 2√2>0

but if you consider case 2: a=8 and b=27
The square root becomes the reverse i.e 2√2 - 3√3...This is negative because 2√2 - 3√3<0

Therefore, the positive square root is 3√3 - 2√2 and it is obtained when a=27 and b=8 in case 1 above

Regards

Thanks bro, God bless you

1 Like

Re: Solve This Mathematics Problem by Johnpaul01: 8:41am On Jan 29, 2017
pls I need solutions to these
1. solve 2x-3y=1,
x²+xy-4y²=2

2. resolve into partial fraction
a. x²/(x+1)²
b. 4x+7/(x²+3)(x²+5)
c. 2x³+3x²+7/(x+1)²(x²+1)

3. When ax³+bx²+cx-4 is divided by x+2. the remainder is double that is obtained when the expression is divided by x+1. show that C can have any value, and find b in term of a.

4. solve x-1/(x-2)(x+1)<0

the mathematicians in d house please help me solve d ones you can
Re: Solve This Mathematics Problem by combine: 1:33pm On Mar 03, 2017
Cheap stuff
Re: Solve This Mathematics Problem by Johnpaul01: 12:55pm On Mar 04, 2017
combine:
Cheap stuff
After we submitted it already??

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