a(2-√3)=b(2-√3)=1, a^2-b^2= ?
Answers
Answer:
0 is the answer
Step-by-step explanation:
thank you
Answer:
before doing anything else, first rationalize the denominator of both a and b to simplify the algebra:
a = (2+sqrt(3))/(4–3) = 2+sqrt(3)
b = (2-sqrt(3))/(4–3) = 2-sqrt(3)
Gaurav Verma
Answered 2 years ago
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^
Chirantan Gupta
Answered 2 years ago
What is the sum of √ (1 + 1/1^2 + 1/2^2) + √ (1 + 1/2^2 + 1/3^2) + … + √ (1 + 1/2007^2 + 1/2008^2)?
First of all your question is utterly wrong. This is not the question. I found it too difficult to figure out what it meant as what you have written does not follow a pattern.Next time post anything after thinking about it first. Anyway Quora is filled up with ego maniacs and imbeciles of which I have personally not a fig of doubt and no interest in.
Anyway here is what your problem is :
1+112+122−−−−−−−−−−√+1+122+132−−−−−−−−−−√+....1+120072+120082−−−−−−−−−−−−−−−−√
Solution:
You see the first term:
1+112+122−−−−−−−−−−√=32=
SusaiRaj
Answered 1 year ago
If a = √8 - √7 and a = 1/b, then what is a^2 + b^2 - 3ab/a^2 + ab + b^2 equal to?
a = √8 - √7 , a = 1/b = b = 1/a = {1/(√8-√7)}
=> b= (√8+√7)/{(√8)^2-(√7)^2} = √8 + √7.
So (a^2+b^2–3ab)/(a^2+ab+b^2)
= {(a-b)^2 - ab}/{a+b)^2-ab}. ……(1)
Now a+b = (√8-√7)+(√8+√7) = 2√8
a - b = (√8-√7)-(√8+√7)= -2√7
ab = (√8-√7)(√8+√7)= 8–7 = 1
Substituting these values in (1) we get
{(-2√7)^2 - 1}/{(2√8)^2–1) = (28–1)/(32–1)
= 27/31
Nischal Joshi
Answered 3 years ago
Where is the mistake in this calculation: -1 = i^2 = √ - 1 * √ - 1 = √(-1)^2 = √1 = 1?
This is a pretty popular fallacy that students tend to learn. And, this is one of the most easily detectable fallacy.
Now without lengthening the answer, I would go straight to the point:
√a∗√b=√a∗b,ifandonlyifa,b>0.
The formula is well recognized by students but they tend to forget or ignore the conditions that the formula is only true if and only if both the numbers are positive.
So, lets see the 3rd and 4th steps in your fallacy,
√-1 * √ -1 = √(-1)(-1) = √(-1)^2
So, I guess you’ve realized the mistake here. The formula you used was valid only if both the numbers were positive, but here -1<0, so the formula ain’t valid and hence the result.
Peter Shea
Answered March 4, 2021
How do I solve √ (1 + ((x^4-16) / (8x^2)) ^2)? Can anyone explain to me?
You don't solve because it's an expression, not an equation. However, you *can* simplify it fairly easily.
Note: I prefer to mix up my parentheses with other kinds of brackets to make it easier to keep track of the pairs.
Let p = √(1 + ((x^4-16) / (8x^2)) ^2)
Now define:
q = x^4-16,
r = 8x^2
Then p = √(1 + [q / r] ^2)
= √([r^2 + q^2] / r^2)
= √[r^2 + q^2] / √[r^2]
= (√[r^2 + q^2] / r)
Substituting the expressions for q and r, then expressing as a rational fraction with denominator (8x^2)^2, you will find that r^2 + q^2 is a perfect square.
The solution is a simple rational fraction very similar to the expression for q / r.
Quora User
Sania Ejaz
Answered 3 years ago
If, A+A^-1=√3 Then what is A^2+A^-2=?
ACCORDING TO THE GIVEN QUESTION
(A+A^-1)=(3)^1/2
A+1/A=(3)^1/2
A=(3)^1/2–1/A
A=((3)^1/2)A-1/A
A/A=((3)^1/2)A-1
1=(3)^1/2A -1
1+1=(3)^1/2A
2/(3)^1/2=A
SO BY PUTTING IN THE GIVING EQUATION
A^2 +A^-2= X
THEN BY PUTTING THE VALUE OF A WE GET :
{2/(3)^1/2}^2 +{2/(3)^1/2}^-2=X
{2/(3)^1/2}^2 +1/{2/(3)^1/2}^2 =X
4/3 +1/4/3=X
4/3 +3/4 =X
OR X=[4*4+3*3]/12
X=[16+9]/12
X=25/12 OR X=2.088
June (Nathan) Richardson
Answered 2 years ago
How would you solve 4 [1/2 (√3+1)] ²+2 [1/2 (√3+1)] ²-8 [1/2 (√3+1)] ²?
Original question: 4 [1/2 (√3+1)] ²+2 [1/2 (√3+1)] ²-8 [1/2 (√3+1)] ²
Let’s rewrite [1/2 (√3+1)] ² as u for simplicity
4u+2u-8u = -2u
Now, resubstitute u to get -2 [1/2 (√3+1)] ²
This can be simplified further:
-2 [1/2 (√3+1)] ²
-2*(1/2)² * -2(√3+1)²
-2*(1/4) * -2*(3+2√3+1)
-1/2 * -2*(4+2√3)
1 * -2(4+2√3)
-8–4√3
We have completely simplified it
Mallapuram Vishnu
Answered 1 year ago
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)
Ruchi Chhabra
Neil Gupta
Answered 2 years ago
What's the value of (3 + 2 √(2) ^1/2 - (3 - 2 √(2) ^1/2)?
=3+22–√12−(3−22–√12)
=3+22–√−−−√−(3−22–√−−−√)
=3+22–√−−−√−3+22–√−−−√
=42–√−−−√
=42–√4
Amisha Kumari
Goh Kim Tee
Answered 1 year ago
What is the value of [1/ (2+√3-2√2)] + [3/ (2+√3+2√2)]?
①N=(2+√3+2√2)+3(2+√3–2√2)
=8+4√3—4√2
=4(2+√3—√2)
②D
=(2+a+2b)(2+a—2b)
=(2+a)²—4b²
=4–4b²+a²+2a
=4–8+3+2√3
=2√3–1
③ N/D
=4(2+√3–√2)(2√3+1)/(12–1)
=4(4√3+6–2√6+2+√3–√2)/11
=4(8+5√3—√2–2√6)/11