Question

The value of \[\sqrt {42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } } \].
A. 7
B. 6
C. 5
D. 3

Answer 1

Hint: This type of question can be solved by considering the given expression as a variable and by taking out the square root by squaring on both sides and then further simplification will get our required result.

Complete step-by-step answer:
Given expression is\[\sqrt {42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } } \],
Now let the given expression be considered as a variable let us take the variable as\[x\],
So the given expression can be written as,
\[\sqrt {42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } } = x\],
Now squaring on both sides we get,
\[{\left( {\sqrt {42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } } } \right)^2} = {x^2}\],
\[ \Rightarrow 42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } = {x^2}\],
Now take 42 on the other side we get,
\[ \Rightarrow \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } = {x^2} - 42\],
Now as we know that as\[\sqrt {42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } } = x\],we get,
\[ \Rightarrow x = {x^2} - 42\],
We now we will make the quadratic equation and solve the equation, so we get,
\[ \Rightarrow {x^2} - x - 42 = 0\],
Now we got the quadratic equation by using quadratic formula which is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]and solve the equation, here \[a = 1,b = - 1\]and\[c = - 42\],
Now substituting the values in the formula we get,
\[ \Rightarrow x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} \pm 4\left( 1 \right)\left( { - 42} \right)} }}{{2\left( 1 \right)}}\],
Now simplifying we get,
\[ \Rightarrow x = \dfrac{{1 \pm \sqrt {1 - \left( { - 168} \right)} }}{2}\],
So again simplifying the root part we get,
\[ \Rightarrow x = \dfrac{{1 \pm \sqrt {169} }}{2}\],
Now removing square root we get,
\[ \Rightarrow x = \dfrac{{1 \pm 13}}{2}\].
As it is having plus and minus sign we will get two values, now calculating we get,
\[ \Rightarrow x = \dfrac{{1 + 13}}{2},x = \dfrac{{1 - 13}}{2}\]
\[ \Rightarrow x = \dfrac{{14}}{2},x = \dfrac{{ - 12}}{2}\],
Now we got the values,
\[ \Rightarrow x = 7,x = - 6\],
So now as the square root cannot have a negative value so, \[x\]cannot be\[ - 6\], so the value of\[x\]is 7.
Now we got\[x = 7\], so the given expression has the value 7.
So,\[\sqrt {42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } } = 7\].

The value of expression\[\sqrt {42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } } \]is 7, i.e., \[\sqrt {42 + \sqrt {42 + \sqrt {42 + \sqrt {42 + ...........\infty } } } } = 7\], so from the options A is the correct option.

Note:
In these types of repeating square questions, we equate the expression to some variable and then squaring on both sides and we will get the quadratic equation, and by using the quadratic formula we can easily find the value of expression.