that makes no sense if you mean square root to the power 2 its equal: the number in the root (let it =x) in absolute value then its equal=: x or -x im not sure
if x is equal to the square root of 81 then x is ±9 (plus or minus 9)
X is greater than 5. If X were equal to 5, X-5 would be 0 which has no real square root. If X were 4, X-5 would be -1 which has no real square root, and so on.
x2 - 6x + 1 = 0 x = [-(-6) +&- square root of [(-6)2 - 4(1)(1)]]/2(1) x = (6 +&- square root of 32)/2 x = [6 +&- 4(square root of 2)]/2 x = 3 +&- 2(square root of 2) x = 3 + 2(square root of 2) or x = 3 - 2(square root of 2) Check:
Square root of 49 = 7... Square root of 9 = 3... 3 x 7 = 21
If x is 0, the square root is 0 also.
Call this number "x". In this case:x = root(5 root(5 root(5 root(5... Since the expression inside the first root is equal to the entire expression, you get: x = root(5 x) Squaring both sides, to get rid of the rood, you get: x squared = 5x x squared - 5x = 0 x(x - 5) = 0 So, x is either equal to 0, or to 5. Indeed, if you start with any number and repeatedly multiply by 5 and then take the square root, you get closer and closer to 5... unless you start with zero, in which case you get exactly zero.
that makes no sense if you mean square root to the power 2 its equal: the number in the root (let it =x) in absolute value then its equal=: x or -x im not sure
if x is equal to the square root of 81 then x is ±9 (plus or minus 9)
X is greater than 5. If X were equal to 5, X-5 would be 0 which has no real square root. If X were 4, X-5 would be -1 which has no real square root, and so on.
No. The square roots of numbers between 0 and 1 (not including 0) are greater than or equal to (in the case of 1) the number. The square root of 0.49 is 0.7 for example.
say x=-2, the x^2=4, but the square root of 4 is 2 because we always take the positive value, known as the principal root. Using this square root of x^2=|x|. So if x is greater than or equal 0, than square root of x^2 is x, but if x is less than zero we must take its abolute value.
0
For example, x0.5 (which is equal to the square root of x).
0 0 x 0 = 0
Domian is x>-6 Range is y> or equal to 0
(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2