The correct option is D 3Given, √(6+√(6+√(6+......∞))) let, x=√(6+......∞) x^2=6+√(6+√(6+√(6+......∞))) x^2=6+x x^2 x 6=0 (x 3)(x+2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Quadratic Formula for Finding Roots

√6+√6+√6+.......

Question

$\sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . . . . . \infty}}}$ is:

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 3
Given,

$\sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . . \infty}}}$

let,

$x = \sqrt{6 + . . . . . . \infty}$

$x^{2} = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . . \infty}}}$

$x^{2} = 6 + x$

$x^{2} - x - 6 = 0$

$(x - 3) (x + 2) = 0$

$x = 3$

Suggest Corrections

Similar questions

If $\frac{6^{6} + 6^{6} + 6^{6} + 6^{6} + 6^{6} + 6^{6}}{3^{6} + 3^{6} + 3^{6}} \div \frac{4^{6} + 4^{6} + 4^{6} + 4^{6}}{2^{6} + 2^{6}} = 2^{n}$ then find the value of $n$ .

Mark the correct option:

The number is equal to

A. 6⁶

\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . .}}}}

up to

\infty

is?

Q. Find the value of

^{6} C_{0} .^{12} C_{6} -^{6} C_{1} .^{11} C_{6} +^{6} C_{2} .^{10} C_{6} +^{6} C_{2} .^{10} C_{6} -^{6} C_{3} .^{9} C_{6} +^{6} C_{4} .^{8} C_{6} -^{6} C_{5} .^{7} C_{6} +^{6} C_{6} .^{6} C_{6}

\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . .}}}}

= ?

Join BYJU'S Learning Program

√6+√6+√6+.........∞ is:

The correct option is D 3Given,√6+√6+√6+......∞let,x=√6+......∞x2=6+√6+√6+√6+......∞x2=6+xx2−x−6=0(x−3)(x+2)=0x=3

$\sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . . . . . \infty}}}$ is:

The correct option is D 3
Given,

$\sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . . \infty}}}$

let,

$x = \sqrt{6 + . . . . . . \infty}$

$x^{2} = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . . \infty}}}$

$x^{2} = 6 + x$

$x^{2} - x - 6 = 0$

$(x - 3) (x + 2) = 0$

$x = 3$