The correct option is A 3 Let x=√(6+√(6+√(6+.......∞))) ⇒ x^2=6+x ⇒ x^2 x 6=0 ⇒ (x 3)(x+2)=0 ⇒ x= 2 #160; or #160; 3 ∴ ...

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Standard XII

Mathematics

Quadratic Formula for Finding Roots

√6+√6+√6+√6+....

Question

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

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Solution

The correct option is A $3$
Let $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . . . \infty}}}$
$\Rightarrow x^{2} = 6 + x$
$\Rightarrow x^{2} - x - 6 = 0$
$\Rightarrow (x - 3) (x + 2) = 0$
$\Rightarrow x = - 2 o r 3$
$∴ x = 3$ only

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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

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√6+√6+√6+√6+.....to ∞..............

The correct option is A 3Let x=√6+√6+√6+.......∞⇒x2=6+x⇒x2−x−6=0⇒(x−3)(x+2)=0⇒x=−2or3∴x=3 only

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

The correct option is A $3$
Let $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . . . \infty}}}$
$\Rightarrow x^{2} = 6 + x$
$\Rightarrow x^{2} - x - 6 = 0$
$\Rightarrow (x - 3) (x + 2) = 0$
$\Rightarrow x = - 2 o r 3$
$∴ x = 3$ only