Question

Solve the following

$\sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } }$

Let assume that

$\rm \: x = \sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } }$

On squaring both sides, we get

$\rm \: {x}^{2} = 12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } }$

$\rm \: {x}^{2} = 12 + x$

$\rm \: {x}^{2} - x - 12 = 0$

Its a quadratic equation, so on splitting the middle terms, we get

$\rm \: {x}^{2} - 4x + 3x - 12 = 0$

$\rm \: x(x - 4) + 3(x - 4) = 0$

$\rm \: (x - 4)(x + 3) = 0$

$\rm\implies \:x = 4 \: \: \: or \: \: \: x = - 3 \: \: \: \{rejected \}$

x = - 3, is rejected because

$\rm \: \sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } } \:>\:0$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \: \sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } } = 4}}$

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ADDITIONAL INFORMATION

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

ANSWER:

4

STEP-BY-STEP EXPLANATION:

Given that $\sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty }}}$

x = $\sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } }$

Let's say $\sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } }$ is x. Since √12 is repeating. So,

x = √(12 + x)

Do squaring on both sides,

→ (x)² = √(12 + x)²

→ x² = 12 + x

→ x² - x - 12 = 0

The above quadratic equation is in the form of ax² + b + c = 0 where a is 1, b is -1 and c is -12. We need to split the middle term in such a way that it's sum is -1 and product is -12.

→ x² - x - 12 = 0

Where -4x + 3x are the perfect splits of the middle term. As it's sum is -x and product is -12.

→ x² + 3x - 4x - 12 = 0

Take x as common from (x² + 3x) and -4 as common from (-4x - 12)

→ x(x + 3) -4(x + 3) = 0

Take (x + 3) as common,

→ (x + 3) (x - 4) = 0

→ x = -3, 4

(neglect the negative value)

Therefore, the value of $\sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty } } }$ is 4.