No.
A surd in the form a√b cannot, in general, be simplified.
If an infinite surd is like (√(6+√(6+√(6+√(6...),to solve it, follow these steps: Set x = √(6+√(6+√(6+√(6... therefore x2 = 6+√(6+√(6+√(6+√(6... therefore x2 = 6+x therefore 6+x-x2 = 0. Factorising the expression gives you (-x+3)(x+2)=0 Only the positive answer need be concerned with: -x+3=0 therefore -x = -3 therefore x = 3.
Normally they are.
yes
2
the same way you find chuck Norris
Yes it is a surd
If the value of the surd is positive, then it will be another surd. Otherwise it will be a complex number.
A surd is the square root of an integer (whole number), e.g. square root of 2 is a surd. Pi is not the square root of an integer so not a surd, just an irrational number
I am pretty sure that the cube root of negative seven is a surd. I checked on the calculator.......and it showed a negative number.....??I think when it is not a surd...it's supposed to say error, so the number probably means that it is a surd..
No. It is not a criminal or civil offence!
No.
i did
A surd in the form a√b cannot, in general, be simplified.
algebra
no