Question

Evaluate 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

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

As,

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

So,

$\rm\implies \:x > 0$

Now,

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

On squaring both sides, we get

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

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

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

Now, its a quadratic equation in x, so using splitting of middle terms, we get

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

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

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

$\rm\implies \:x = 4 \: \: or \: \: x = - 3 \: \red{\{rejected \: as \: x > 0 \}}$

Hence,

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

$\rule{190pt}{2pt}$

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:

Question :-

★ Evaluate the following :-

$\mapsto \sf \sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty}}}$

Given :-

$\mapsto \sf \sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty}}}$

To Find :-

$\mapsto$ Evaluate.

Solution :-

Let,

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

So,

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

By squaring both sides we get :

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

$\implies \sf 12 + x =\: x^2$

$\implies \sf x + 12 =\: x^2$

$\implies \sf - x^2 + x + 12 =\: 0$

$\implies \sf - (x^2 - x - 12) =\: 0$

$\implies \sf x^2 - x - 12 =\: 0$

By doing middle term break we get :

$\implies \sf x^2 - (4 - 3)x - 12 =\: 0$

$\implies \sf x^2 - 4x + 3x - 12 =\: 0\: \: \small\bigg\lgroup \sf\bold{\pink{By\: doing\: middle-term\: break}}\bigg\rgroup$

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

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

$\implies \bf (x - 4) =\: 0$

$\implies \sf x - 4 =\: 0$

$\implies \sf\bold{\purple{x =\: 4}}$

Or,

$\implies \bf (x + 3) =\: 0$

$\implies \sf x + 3 =\: 0$

$\implies \sf\bold{\purple{x =\: - 3}}\: \: \small\bigg\lgroup \sf\bold{\pink{Number\: can't\: be\: negetive}}\bigg\rgroup$

So, the value of x = 4.

Hence,

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

$\dashrightarrow \sf\bold{\red{\sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty}}} =\: 4}}$

$\sf\boxed{\bold{\therefore\: \sqrt{12 + \sqrt{12 + \sqrt{12 + - - - \infty}}} =\: 4\: .}}$