if a=1/2-under root 3,b=1/2+under root 3,then find the value of (a+b/a-b)^2
Answers
Answer:
a = 1/(2–sq rt 3)
b = 1/(2+sq rt 3)
Question is what is a^2 + b^2
As we know (a+b)^2 = a^2 + b^2 + 2ab
So, a^2 + b^2 = (a+b)^2 - 2ab
Replace a & b to RHS of above equation
a^2 + b^2 = (1/2–sq rt 3+1/(2+sq rt 3))^2 - 2 * 1/(2–sq rt 3) * 1/(2+sq rt 3)
= ((2+sq rt 3+ 2–sq rt 3)/(2–sq rt 3)(2+sq rt 3))^2 - 2 * 1/(2–sq rt 3)(2+sq rt 3)
=(4/(4–3))^2 - 2 * (1/4–3)
= 4^2 - 2 * 1/1
= 16–2
So, a^2 + b^2 = 14
If a= (√3+√2) / (√3-√2) and b= (√3-√2) / (√3+√2), then how do I find a^2+b^2?
What is the value of [1/ (2+√3-2√2)] + [3/ (2+√3+2√2)]?
What is (2+√3) / (2-√3) + (2-√3) /(2+√3) + (√3-1)/(√3+1)?
If 1/b=2+√3, what is the value of b^4+1/b^4?
If x=3+√8, then how do I find the value of (x²+1/x²)?
Question: Given that a = 1/(2-√3) and b = 1/(2+ √3), how do I calculate a^2 + b^2?
Well this is a rather simple question
You need the concept of rationalization as we can’t have imaginary terms in the denominator.
So using this concept and that we have the values of a and b we can easily find a^2+b^2.
And so here is the solution.
a=12−3–√=12−3–√⋅2+3–√2+3–√=2+3–√
b=12+3–√=12+3–√⋅2−3–√2−3–√=2−3–√
a2+b2=(2−3–√)2+(2+3–√)2=4−43–√+3+4+43–√+3=14
We start by making the denomination terms as a simpler number i.e., in the first case
2-√3
2-1.732=-0.268
I'm the second case we need to add the terms i.e.,
2+√3
2+1.732=3.732
Now we substitute these answers in the equation.
a=1/(-0.268)=-3.7313
Then, b=1/3.732=0.2679
according to the equation,
a^2+b^2={(-3.7313)^2}+{(0.2679)^2}
(13.9229)+(0.07179)=13.99469
This is the solution to the problem.
What's the value of (3 + 2 √(2) ^1/2 - (3 - 2 √(2) ^1/2)?
If a = 1/ (2-√3) and b=1/ (2+√3), then what is the value of 7a² +11ab?
What is? C^2=√ (a^2+b^2)
What is the sum of 11+22+33+⋯+nn ?
If x=2+2*2/3+2*1/3 what is the value of x^3-6x^2+6x?
3 things to know:
1 (1x)2=1x2
2 (a±b)2=a2±2ab+b2
3 ab±cd=ad±bcbd
So we can conclude that (12−3√)2+(12+3√)2=14 .
Read Bruno’s answer. He almost did it. I will finish it.
a^2=1/(2-sqrt(3)(^2 = 1/(4–4sqrt(3)+3) = 1/(7–4sqrt(3))
b^2 = 1/(7+4sqrt(3))
Now we add them. We are adding two fractions, so we need to find their common denominator. It is (7–4sqrt(3))*(7+4sqrt(3)) = 49–16*3 = 1
a^2+b^2 = (7+4sqrt(3) + 7-4sqrt(3))/1 = 14/1 = 14
I’ll upload the explanation on how I solve this question.
About the last part of the question, I am not sure there’s a way to solve this. That’s all I could do about this question. Hope you could understand my explanation. Don’t hesitate to ask If you have any questions.
a^2 + b^2 = (a+b)^2 -2ab
Also
1/(x+y) + 1/(x-y) = (x-y + x+y)/(x^2-y^2)
For x=2 y=sqrt (3) , the above equals 4.
And
1/(x+y) + 1/(x-y) = 1/(x^2-y^2) = 1/1 =1 For x=2 y=sqrt (3) .
So a^2+b^2 = 4^2 -2 = 14
If a= (√3+√2) / (√3-√2) and b= (√3-√2) / (√3+√2), then how do I find a^2+b^2?
This problem can be solved in 2 ways.
1st :-
a=3–√+2–√3–√−2–√
=(3–√+2–√)(3–√+2–√)(3–√−2–√)(3–√+2–√)
=(3–√+2–√)23−2
=((3–√)2+(2–√)2+2(3–√)(2–√)
=3+2+26–√
=5+26–√
b=3–√−2–√3–√+2–√
=(3–√−2–√)(3–√−2–√)(3–√+2–√)(3–√−2–√)
=(3–√−2–√)23−2
=((3–√)2+(2–√)2+−2(3–√)(2–√)
=3+2−26–√
=5−26–√
Now
a2+b2
We know that
(x+y)^2= x^
If ‘x = a+√a²+1’, then what is ‘a’ in terms of ‘x’?
x = a + √a^2 +1
1/x = 1/(a + √a^2 + 1 )
Multiplying and dividing R.H.S. by
(a - √(a^2+1) ) , we get
1/x = (a - √(a^2 +1) )/(a^2 -(a^2 +1) )
=> 1/x = -(a - √(a^2 +1) ) = -a + √(a^2 +1)
Subtracting 1/x from x we get
x - 1/x = 2a
=> a = (x - 1/x )/2
Hope it helps !!!
If a = 1/ (2-√3) and b=1/ (2+√3), then what is the value of 7a² +11ab?
Clearly ab= 1/(2^2–3) = 1. So ab=1
By rationalizing we get a= 2+3^(1/2) —»a^2= 4+3+4*3^(1/2)= 7+4*3^(1/2)
-By Substituting a^2 and ab values in 7a^2+11ab we have 7a^2+11ab= 7(7+4*3^(1/2))+11= 49+28*3^(1/2)+11 = 60+28*3^(1/2)
If a= (√3+√2) / (√3-√2) and b= (√3-√2) / (√3+√2), then how do I find a^2+b^2?
What is the value of [1/ (2+√3-2√2)] + [3/ (2+√3+2√2)]?
What is (2+√3) / (2-√3) + (2-√3) /(2+√3) + (√3-1)/(√3+1)?
If 1/b=2+√3, what is the value of b^4+1/b^4?
If x=3+√8, then how do I find the value of (x²+1/x²)?
What's the value of (3 + 2 √(2) ^1/2 - (3 - 2 √(2) ^1/2)?
If a = 1/ (2-√3) and b=1/ (2+√3), then what is the value of 7a² +11ab?
What is? C^2=√ (a^2+b^2)
What is the sum of 11+22+33+⋯+nn ?
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