Under root x+ under root x up to infinite time =6 find x
√(x+√(x+√(x+...))) = 6
square both sides:
x+√(x+√(x+√(x+...))) = 36
x = 36-√(x+√(x+√(x+...)))
x = 36-6 = 30
Loda answer h
To find the value of x in the equation √x + √x + √x + ... (infinity times) = 6, we need to think about the pattern being repeated.
Let's consider a single repetition of the pattern:
√x + √x + √x + ... (repeated n times).
This can be simplified as n * √x.
Applying this logic to the original equation, we can rewrite it as:
√x + √x + √x + ... (infinity times) = 6
∞ * √x = 6
Now, we will solve for √x:
√x = 6/∞
As the denominator approaches infinity, the value of 6/∞ becomes infinitesimally small, resulting in √x being equal to 0.
Now, squaring both sides to isolate x:
x = (√x)^2
x = 0^2
x = 0
Therefore, the value of x in the equation is 0.
To solve the equation √x + √x + √x + ... = 6 (with the square root repeated infinitely), we can use a method called algebraic manipulation.
Let's denote the value of √x as a variable, say y.
So, we can rewrite the equation as y + y + y + ... = 6.
Now, since we have an infinite number of y terms on the left side, we can express it as y times infinity: y * ∞.
However, infinity is not a number but a concept, so we cannot perform ordinary algebraic operations with it. To handle this, we transform our equation by multiplying both sides by y:
y + y + y + ... = 6
y * ∞ = 6y
Now, we have the equation y * ∞ = 6y, where y ≠ 0.
To simplify further, we divide both sides of the equation by y:
y * ∞ / y = 6y / y
∞ = 6
This equation shows that infinity is equal to 6. However, this is not a valid mathematical result. The equation you provided leads to a contradiction because infinity cannot be treated as a finite number.
Therefore, there is no real value of x that satisfies the given equation.