The square root of 2 is irrational, and therefore, this equation is unsolvable.
Square root of 10 is irrational.
sqrt(8) = sqrt(4*2) = 2*sqrt(2).Even without given that sqrt(2) is a rational, you can give that the square root of 2 starts converging onto the "Pythagoras Constant" eventually, as it takes an infinite amount of digits to square root an integer that is not perfectly squared.Thus, an rational x irrational = irrational, thus the sqrt(8) is irrational (an approximation is 2.8284271247...).
The square root of 128 is 8 times the square root of 2 because it's an irrational number. An estimate is 11.313708498984760390413509793678
It is known that the square root of an integer is either an integer or irrational. If we square root2 root3 we get 6. The square root of 6 is irrational. Therefore, root2 root3 is irrational.
it is the square root of 16 which equals 4.
3.6
It is an irrational number
irrational
No it's 8, which is not irrational.
The square root of 8 is irrational and real.
The square root of 8 is an irrational number because it cannot be represented as a fraction of two integers.
The square root of 8 over 25 is irrational, and real.
It can't be expressed as a whole number. The square root of two is irrational, and the product of an irrational number and a rational number is also irrational. So 8*sqrt(2) is irrational.
yes because √2 is irrational and √8= √[4*2]=2√2
The square root of 64 is 8 which is a rational number
Yes the result would be an irrational number.