Question

The value of $\sqrt {7 + \sqrt {7 - \sqrt {7 + \sqrt {7 - .....\infty } } } } $ is
A) $5$
B) $2$
C) $3$
D) $4$

Answer 1

Hint: Infinity is used to explain events that are boundless and limitless . If given any function which goes up to infinity then we take it as a variable. After taking it we arrange them and we cut small parts from its beginning then it cannot change its character. As example we take \[y = \sqrt {a + \sqrt {a + \sqrt {a + \sqrt {a + .... + \infty } } } } \] , if we cut first one term then the variable cannot change . We also write this as $y = \sqrt {a + y} $ and solve it to get the value .

Complete step by step answer:
First we take the given data i.e., $\sqrt {7 + \sqrt {7 - \sqrt {7 + \sqrt {7 - .....\infty } } } } $
Find the value of this function one by one , that is not possible so we can take $x = \sqrt {7 + \sqrt {7 - \sqrt {7 + \sqrt {7 - .....\infty } } } } $
Now $x = \sqrt {7 + \sqrt {7 - \sqrt {7 + \sqrt {7 - .....\infty } } } } $
Since $x = \sqrt {7 + \sqrt {7 - \sqrt {7 + \sqrt {7 - .....\infty } } } } $ then if we put this after two term that again same because this function goes up to infinity .
$ \Rightarrow x = \sqrt {7 + \sqrt {7 - x} } $
Squaring both sides of the above equation and we get
$ \Rightarrow {x^2} = 7 + \sqrt {7 - x} $
$ \Rightarrow {x^2} - 7 = \sqrt {7 - x} $
Again take square both sides of the above equation and we get
$ \Rightarrow {\left( {{x^2} - 7} \right)^2} = 7 - x$
Use the formula of ${(a - b)^2} = {a^2} - 2ab + {b^2}$ in the above equation and we get
$ \Rightarrow {x^4} - 14{x^2} + 49 - 7 + x = 0$
$ \Rightarrow {x^4} - 14{x^2} + x + 42 = 0$
Use vanishing method in above equation and we get
$ \Rightarrow {x^3}(x - 3) + 3{x^2}(x - 3) - 5x(x - 3) - 14(x - 3) = 0$
$ \Rightarrow (x - 3)({x^3} + 3{x^2} - 5x - 14) = 0$
Again use vanishing method in the above equation and we get
$ \Rightarrow (x - 3)\{ {x^2}(x + 2) + x(x + 2) - 7(x + 2)\} = 0$
$ \Rightarrow (x - 3)(x + 2)({x^2} + x - 7) = 0$
If any product of some function or polynomial equal to zero then we know any one of them equal zero
Use this in the above equation and we get
$ \Rightarrow x - 3 = 0$ or $x + 2 = 0$ or ${x^2} + x - 7 = 0$
Use the formula of Sridhar acharya in above quadratic formula and we get
$ \Rightarrow x = 3$ or $x = - 2$ or $x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 1 \times ( - 7)} }}{{2 \times 1}}$
$ \Rightarrow x = 3$ or $x = - 2$ or $x = \dfrac{{ - 1 \pm \sqrt {29} }}{2}$
$ \Rightarrow x = 3$ or $x = - 2$ or $x = \dfrac{{ - 1 + \sqrt {29} }}{2},\dfrac{{ - 1 - \sqrt {29} }}{2}$
We take $x = \sqrt {7 + \sqrt {7 - \sqrt {7 + \sqrt {7 - .....\infty } } } } $ , then we know the value of $x$ is always greater than $\sqrt 7 $ i.e., $x > \sqrt 7 $ .
From above condition we get the value of $x = 3$
$\therefore $ $x=3$. So, Option (C) is correct .

Note:
Formula of Shridhar Acharya for quadratic equation is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , where the quadratic equation is $a{x^2} + bx + c = 0$ . It is also known as the quadratic equation . Vanishing method is a popular method to solve the higher degree polynomial . A constant value like $a$ which satisfies the equation then we take common $(x - a)$ from the given polynomial.