Question

please solve it now it's my homework

Answer 1

Let

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


Now square both the sides

$= > ( \sqrt{6 + \sqrt{6 + \sqrt{6....} } } ) {}^{2} = x {}^{2} \\ \\ = > 6 + \sqrt{6 + \sqrt{6 + \sqrt{6...} } } = {x}^{2}$


But we have assumed that

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

Hence,
$6 + \sqrt{6 + \sqrt{6 + \sqrt{6.....} } } = {x}^{2} \\ \\ can \: be \: written \: as \: \\ \\ 6 + x = {x}^{2}$

Now solve the quadratic equation

${x}^{2} = 6 + x \\ \\ = > {x}^{2} - x - 6 = 0 \\ \\ = > {x}^{2} - 3x + 2x - 6 = 0 \\ \\ = > x(x - 3) + 2(x - 3) = 0 \\ \\ = > (x + 2)(x - 3) = 0$

Now,

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

=> (x + 2) = 0/(x - 3)

=> x + 2 = 0

=> x = -2


Similarly,

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

=> x - 3 = 0

=> x = 3

So x = -2, 3

We generally take the positive greater value.

=> x = 3


Answer :- 3

Answer 2

ĀNSWĒR ⬆⬆


THANKS ✌☺


#DILDAAR NAVI ♥