Question

Find the Value of
$\sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + .... \infty } } } } = \: ?$
​

Answer 1

AnswEr :

• Let's Know the Rule First of Solving Question like this :

$\mathsf{\sqrt{a + \sqrt{a + \sqrt{a + \sqrt{a + ... \infty } } } } }$

We Will Break a into the form of {n × (n + 1)}

⋆ And the Answer will be (n + 1) in the situation of Addition.

• Just like that we can follow this too :

$\mathsf{\sqrt{a - \sqrt{a - \sqrt{a - \sqrt{a - ... \infty } } } } }$

We Will Break a into the form of {n × (n + 1)}

⋆ And the Answer will be n in the situation of Subtraction.

• Now Let's Head to the Question :

$\mathsf{\sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + ... \infty } } } } }$

• We Can Break 12 in the Form of {n × (n + 1)} as {3 × (3 + 1)}
• n = 3 and, (n + 1) = (3 + 1) = 4

⋆ As we will take (n + 1) as Answer. That's why Required Answer will be 4.

$\mathsf{\sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + ... \infty } } } } = 4 }$

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⋆ If You Want to Solve this By Complete Method then here You Go :

Let's Assume x is Equal to ;

$\longrightarrow\mathsf{x = \sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + ... \infty } } } } }$

• Squaring Both Sides

$\longrightarrow\mathsf{ {(x)}^{2} = \bigg(\sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + ... \infty } } } } \bigg)^{2} }$

$\longrightarrow\mathsf{ {x}^{2} = 12 + {\sqrt{12 + \sqrt{12 + \sqrt{12 + ... \infty } } } } }$

$\longrightarrow\mathsf{ {x}^{2} = 12 +x}$

• As we have taken Value of $\mathsf{x = \sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + ... \infty } } } } }$

$\longrightarrow\mathsf{ {x}^{2} - x - 12 = 0 }$

$\longrightarrow\mathsf{ {x}^{2} - 4x + 3x- 12 = 0 }$

$\longrightarrow\mathsf{ x(x - 4) + 3(x - 4) = 0 }$

$\longrightarrow\mathsf{(x - 4)(x + 3) = 0}$

$\longrightarrow\mathsf{ x = 4 \: \: or \: \: x = - 3}$

⋆ we will Ignore the Value of x = - 3

$\longrightarrow\mathsf{\sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + ... \infty } } } } = 4 }$

჻ Therefore, Value will be 4.