If x = sqrt{1 + sqrt{1 + sqrt{1 + dots infty} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$.Since the expression is infinite,we can write $x = \sqrt{1 + x}$.Squaring both sides,we

Vedclass · Accepted Answer

$\frac{1 + \sqrt{5}}{2}$

Vedclass · Answer

(A) Given $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$.Since the expression is infinite,we can write $x = \sqrt{1 + x}$.Squaring both sides,we get $x^2 = 1 + x$.Rearranging the terms,we obtain the quadratic equation $x^2 - x - 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 1, b = -1, c = -1$:$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$.Since $x$ represents a square root,$x$ must be positive $(x > 0)$.Therefore,we discard the negative root $\frac{1 - \sqrt{5}}{2}$.Thus,$x = \frac{1 + \sqrt{5}}{2}$.

If $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$,then $x =$

Similar Questions

If the set of all $a \in R$,for which the equation $2x^2 + (a-5)x + (15-3a) = 0$ has no real root,is the interval $(\alpha, \beta)$,and $X = \{x \in Z : \alpha < x < \beta\}$,then $\sum_{x \in X} x^2$ is equal to

The number of real solutions of the equation $x|x+5|+2|x+7|-2=0$ is .....................

The number of distinct real roots of the equation $|x||x+2|-5|x+1|-1=0$ is:

Solve the equation $x^{2}-x+2=0$.

The roots $\alpha, \beta$ of the equation $x^2-6(k-1)x+4(k-2)=0$ are equal in magnitude but opposite in sign. If $\alpha > \beta$,then the product of the roots of the equation $2x^2-\alpha x+6\beta(\alpha+1)=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module