Property of Infinites

$x=\sqrt{90+\sqrt{90+\sqrt{90+...}}}$

$\Rightarrow x^{2}=90+\sqrt{90+\sqrt{90+\sqrt{90+...}}}$

$\Rightarrow x^{2}=90+x$

$\Rightarrow x^{2}-x-90=0$

$\Rightarrow (x+9)(x-10)=0$

$x=-9,10$

Since the problem was expressed is positive terms...

$\boxed{x=10}$

$\begin{aligned} x &= \sqrt{ 90 + \sqrt{90 + \sqrt{90 + ...........}}} \\ \text {squaring both sides} \\ x^2 &= 90 + \sqrt{90 + \sqrt {90 + \cdots }} = 90 + x \\ \end{aligned}$

$x^2 - x - 90 =0 \\ (x -10)(x+9) = 0 \\ x = 10 \\$

x = (√90+(√90+(√90...)))

x^2 = 90+(√90+(√90+(√90...)))

So x^2 = 90 + x

=> x^2 - x - 90 = 0

=> x=-9 or x=10

We take x = 10 since the square root must be positive.

so x = 10

x=sqrt{90+x} hence solve

Let $x=\sqrt{90+\sqrt{90+\sqrt{90+\dotsm}}}$ $x=\sqrt{90+x}$ $x^2=90+x$ $x^2-x-90=0$ $x=10 or x=-9$ as the infinite nested radical is positive so we discard the negative root and so $x=10$

Let the given term be 'I' Then I=√(90+I) bcoz its an infinte series Then square both sides and solve the quadratic so formed

We square both sides. Since √(90+√(90+√(90+√(90+...)))) is till infinity, we basically have x²=90+x solving which we get x=-9 or 10. Since we have the √ sign, we consider the positive answer. -9 would have also been correct if the question was x=(90+(90+(90+...)^½)^½)^½

Let x=√(90+√(90+ ...))

Therefore, we can say that x=√(90+x) (Since the expression keeps going on)

Square both sides

x^2=90 + x

x^2 - x - 90 = 0

(x-10)(x+9)= 0

If the product of two terms is 0, at least one of the terms must be zero. Therefore either x-10=0, meaning x=10 or x+9=0, meaning x= -9

However the √ symbol means we will only consider the positive radical, so x ≥ 0.

Clearly, out of our two solutions, x= 10 is the only solution that fits the requirement. Therefore, the value of the expression is 10.