Yes, it is. sqrt(a+b)=sqrt(b+a) sqrt(a) times sqrt(b) = sqrt(b) times sqrt(a)
Assuming the equation is sqrt (12/b2), the solution is: sqrt 12 / sqrt b2 = sqrt (4*3) / b = 2 sqrt 3 / b
Geometric Mean of A and B is the square root of A times B, in this case sqrt 1600 which is 40
Okay, this is the rule for dividing surds: sqrt (a) ______ sqrt (b) = sqrt (a/b) so for example you had sqrt 3 _____ sqrt 21 = sqrt (3/21) = sqrt (7) more complicated... 3 sqrt(3) ________ 4 sqrt(21) = 3/4 sqrt (3/21) = 3/4 sqrt (7) It's pretty easy as long as you can remember the rule. I hope that helps. XD
Suppose A = 2 + sqrt(3) and B = 5 - sqrt(3) Then A and B are two irrational numbers but A + B = 2 + sqrt(3) + 5 - sqrt(3) = 7 which is rational.
Yes, it is. sqrt(a+b)=sqrt(b+a) sqrt(a) times sqrt(b) = sqrt(b) times sqrt(a)
The question is ambiguous because it could refer to [sqrt(A2B) + sqrt(AB2)]/sqrt(AB) = [A*sqrt(B) + B*sqrt(A)]/[sqrt(A)*sqrt(B)] = A/sqrt(A) + B/sqrt(B) = sqrt(A) + sqrt(B) or sqrt(A2B) + sqrt(AB2)/sqrt(AB) = A*sqrt(B) + B*sqrt(A)/[sqrt(A)*sqrt(B)] = A*sqrt(B) + B/sqrt(B) = A*sqrt(B) + sqrt(B) = sqrt(B)*(1 + A)
If there is only the radical, sqrt(b), in the denominator, the form of the fraction is sqrt(b)/sqrt(b).If the denominator is of the form a + sqrt(b) then the form of the fraction is [a - sqrt(b)]/[a - sqrt(b)].It is also possible to use [-a + sqrt(b)]/[-a + sqrt(b)], and this form may be preferred is a^2 < b.
sqrt(a)+sqrt(b) is different from sqrt(a+b) unless a=0 and/or b=0. *sqrt=square root of
Assuming the equation is sqrt (12/b2), the solution is: sqrt 12 / sqrt b2 = sqrt (4*3) / b = 2 sqrt 3 / b
3
sqrt[(a + b)2*(c + d)/pi] = (a + b)*sqrt[(c + d)/pi]
No. The product of sqrt(2) and sqrt(2) is 2, a rational number. Consider surds of the form a+sqrt(b) where a and b are rational but sqrt(b) is irrational. The surd has a conjugate pair which is a - sqrt(b). Both these are irrational, but their product is a2 - b, which is rational.
Notice that 720 = (4 x 5 x 36)sqrt(720) = sqrt(4 times 5 times 36)= sqrt(4) times (sqrt(5) times sqrt(36)= (2) times sqrt(5) times (6)= 12 sqrt(5)
sqrt( x13 ) = sqrt ( x12 times x ) = sqrt ( x12 ) times sqrt( x ) = x6 times sqrt( x ).
If a and b are rational, with a < b, then a + (b-a) [sqrt(2)/ 2] is an irrational number between a and b. This number is between a and b because sqrt(2)/2 is less than one and positive, so that a < a + (b-a) [sqrt(2)/3] < a + (b-a) [1] = b. To prove that a + (b-a) [sqrt(2)/2] is not rational, suppose that a + (b-a) [sqrt(2)/2] = p/q where p and q are integers. Then, sqrt(2) = ( p/q -a ) 2/(b-a) which is rational since the rationals are a field, closed under arithmetical operation, but sqrt(2) not rational
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