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.
Square root of 49 = 7... Square root of 9 = 3... 3 x 7 = 21
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:
If x is 0, the square root is 0 also.
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)
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.
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.
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
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.