The correct option is C 3 Let x= √(6+ √(6+√(6+√(6+...∞)))) #10; #10;On squaring both sides, #10; #10; x^2= 6+√(6+ #10;√(6+√(6+√(6+.. ...

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

Special Products

√ 6+√ 6+√ 6+√...

Question

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . t o \infty}}}} =$ ........

3

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

2

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

1

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

\pm 3

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

Open in App

Solution

The correct option is C $3$
$L e t x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . \infty}}}}$

On squaring both sides,

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

$x^{2} = 6 + x$

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

$x = \frac{1 \pm \sqrt{1^{2} - 4 \times 1 \times (- 6)}}{2 (1)}$

$x = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2}$

$∴ x = 3$ or $x = - 2$

Since a negative number cannot be under root, hence $x = 3$

Suggest Corrections

Similar questions

Question 4(d) :

Look at the figures and write ‘<’ or ‘>’ between the given pairs of fractions:

$\frac{6}{6} □ \frac{3}{3}$

lim n \to \infty \frac{(1^{2} + 2^{2} + 3^{2} + . . . + n^{2}) (1^{3} + 2^{3} + 3^{3} + . . . + n^{3})}{1^{6} + 2^{6} + 3^{6} + . . . + n^{6}}

1^{6} - 2^{6} + 3^{6} - . . . . + (- 1)^{n - 1} n^{6} = \frac{(- 1)^{n - 1}}{2} (n^{6} + 3 n^{5} - 5 n^{3} + 3 n)

Given a = 6 + √3
⇒ (1/a) = 1 / (6 + √3)
Multiply and divide by (6 − √3)
⇒ (1/a) = (6− √3) / [(6 + √3)(6 − √3)]
⇒ (1/a) = (6 − √3) / [36 − 3]
⇒ (1/a) = (6− √3)/33
∴ a - (1/a) = (6 + √3) -(6 − √3)/33
= [33(6 + √3) - (6 - √3)]/33
= [32*6 + 34*√3]/33
= [192 + 34√3]/33

Answer : [192 + 34√3]/33

Why does 1/a come above

Q. How many 6s are there in the following number series, each of which is immediately preceded by 1 or 5 and immediately followed by 3 or 9?
2 6 3 7 5 6 4 2 9 6 1 3 4 1 6 3 9 1 5 6 9 2 3 1 6 5 4 3 2 1 9 6 7 1 6 3

Join BYJU'S Learning Program

√6+√6+√6+√6+.....to∞=........

The correct option is C 3Letx=√6+√6+√6+√6+...∞ On squaring both sides, x2=6+√6+√6+√6+√6+...∞ x2=6+x x2−x−6=0 x=1±√12−4×1×(−6)2(1) x=1±√252=1±52 ∴x=3 or x=−2 Since a negative number cannot be under root, hence x=3

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . t o \infty}}}} =$ ........