Question

Find the value of x if x = root 6 + root 6 + root 6 + Infinity. Where x is a natural number.

Answer 1

Hi there!

Given : x = √6 + √6 + √6 +... ---(i)

Then,
x = √6 + x --- From Eqn. (i)

On squaring both sides,

⇒ x² = 6 + x

⇒ x² - x - 6 = 0

⇒ x² - 3x + 2x - 6 = 0

⇒ x(x-3) + 2(x-3) = 0

⇒ x = 3 Or x = -2

But x is a natural number.
∴ x = 3

[ Thank you! for asking the question. ]
Hope it helps!

Answer 2

Given:

$x=\sqrt{6+\sqrt{6+\sqrt{6+...} } }$where x is a natural number

To Find:

Find the value of x

Solution:

To find these kinds of problems which has no ending or the equations till infinity then we need to find a pattern in the infinity equation and then put that value as what it used to be equal to then it can be easily solved. Let us understand this by solving this question

$x=\sqrt{6+\sqrt{6+\sqrt{6+...} } }$

as we can observe and say that the equation after the first term that is from the 2nd term till infinity is also equal to x. so,

$x=\sqrt{6+\sqrt{6+\sqrt{6+...} } } \\x=\sqrt{6+x} \\x^2-x-6=0$

solving this quadratic equation using the quadratic formula

$x=\frac{1\pm \sqrt{1+24} }{2} \\=3 or -2$

we know that the value of the square root of any number cannot be negative so considering the positive value x that is x=3

Hence, the value of x=3.