Wiki User
∙ 14y agoI don't think you wanted the word "radical" in your question. Aren't you working with: C2 = A2 + B2 ? Or maybe: C = sqrt ( A2 + B2 ) ? In either case, A = sqrt ( C2 - B2 ). If your question is really the way you typed it, then the answer is more complicated.
Wiki User
∙ 14y agoIt is sqrt(0), which equals 0.
The square root of six divided by the square root of two equals the square root of three.
Basically, each number is the square root of its square. So, if you square 59, the result will tell you what number you need to take the square root of, to get 59.
If the product of two irrationals is a rational, then they are both the same radical of a non-perfect square. For example, radical 5 times radical 5 is 5, since that is by definiton what a radical is.
The square root of 5 radical 68 simplifies to 34 radical 5.
It is sqrt(0), which equals 0.
The square root of 136 would be broken up into radical 4 and radical 34, because 4 is the highest perfect square that goes into 136 evenly, and radical 4 equals 2. So, your final answer would be 2 radical 34.
Since 60 can be factored to 4 times 15, then radical 60 equals 2 times radical 15.
The square root of six divided by the square root of two equals the square root of three.
Basically, each number is the square root of its square. So, if you square 59, the result will tell you what number you need to take the square root of, to get 59.
The square-root of B2 is +B and also -B .
There is no "radical square root". Radical means the same as root, it may specifically refer to the square root.
You can factor out the Root(16) which equals 4. So the answer is 4 Root(2).
You can factor out the Root(16) which equals 4. So the answer is 4 Root(3).
If the product of two irrationals is a rational, then they are both the same radical of a non-perfect square. For example, radical 5 times radical 5 is 5, since that is by definiton what a radical is.
If the radical is the square root of a quantity, then yes.
The square root of 5 radical 68 simplifies to 34 radical 5.