The answer depends on the value of x. The value of x must also be >= 0 in order for the function to be valid in the real number system.
The square root of 5x can be written as:
sqrt (5x) or (5x)^(1/2) or (5x) ^.5
Below are a few examples:
sqrt (5*1) = sqrt 5 = ~2.236 (~ means approximately)
(5*1)^(1/2) = 5^.5 = ~2.236 (~ means approximately)
(5*1)^(.) = 5^.5 = ~2.236 (~ means approximately)
sqrt (5*5) = sqrt 25 = 5
5x = sqrt(10) + 15x -10x = sqrt(10) x = -sqrt(10)/10 = -sqrt(10)
-2 and -3Check:(-2) + (-3) = -5(-2)(-3) = 6Thus -2 and -3 are not the required numbers. let's find them: x + y = -6xy = -5 y = -x -6x(-x - 6) = -5-x^2 - 6x = -5x^2 + 6x = 5x^2 + 6x + 9 = 5 + 9(x + 3)^2 = 14x + 3 = (+ & -)square root of 14x = -3 (+ & -)square root of 14x = -3 + square root of 14 or x = - 3 - square root of 14y = -x - 6y = 3 - square root of 14 - 6 or y = 3 + square root of 14 - 6y = -3 - square root of 14 or y = -3 + square root of 14Check:(-3 + square root of 14) + (-3 - square root of 14) = -6(-3 + square root of 14)(-3 - square root of 14) = -5 ?(-3)^2 - (square root of 14)^2 = -5 ?9 - 14 = -5Check also tow other numbers.
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The cube root of 125 is 5 -- that is 5*5*5 = 125
square root 2 times square root 3 times square root 8
a+ square root of b has a conjugate a- square root of b and this is used rationalize the denominator when it contains a square root. If we want to multiply 5 x square root of 10 by something to get rid of the radical you can multiply it by square root of 10. But if we look at 5x( square root of 10 as ) 0+ 5x square root of 10 then the conjugate would be -5x square root of 10
5x times the square root of y
The square root of anything to an even power is that thing to half the power. √(5x^13) = √(x^12 5x) = √(x^12) √(5x) = x^6 √(5x) But is that really any simpler?
5x
Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)
(5x - i)(5x + i) where i is the imaginary unit equivalent to the square root of negative one.
(5x - i)(5x + i) where i is the imaginary unit equivalent to the square root of negative one.
If you mean: 2x^2 -5x = 35 then 2x^2 -5x -35 = 0 and by using the quadratic equation formula the value of x is (5-square root 305)/4 or (5+square root 305)/4
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.
5x = sqrt(10) + 15x -10x = sqrt(10) x = -sqrt(10)/10 = -sqrt(10)
-2 and -3Check:(-2) + (-3) = -5(-2)(-3) = 6Thus -2 and -3 are not the required numbers. let's find them: x + y = -6xy = -5 y = -x -6x(-x - 6) = -5-x^2 - 6x = -5x^2 + 6x = 5x^2 + 6x + 9 = 5 + 9(x + 3)^2 = 14x + 3 = (+ & -)square root of 14x = -3 (+ & -)square root of 14x = -3 + square root of 14 or x = - 3 - square root of 14y = -x - 6y = 3 - square root of 14 - 6 or y = 3 + square root of 14 - 6y = -3 - square root of 14 or y = -3 + square root of 14Check:(-3 + square root of 14) + (-3 - square root of 14) = -6(-3 + square root of 14)(-3 - square root of 14) = -5 ?(-3)^2 - (square root of 14)^2 = -5 ?9 - 14 = -5Check also tow other numbers.
It is 5*x^3*y*sqrt(2x)