Question

Evaluate this question please..

Answer 1

Answer :-

The answer is c) (√13 + 3)/2.

Solution :-

$\sf x = \sqrt{10 + 3 \times \sqrt{10 - 3 \times \sqrt{10 + 3 \times \sqrt{10 - ....... \infty} } } }$

$\sf \implies x = \sqrt{10 + 3 \times \sqrt{10 - 3 \times \bigg( \sqrt{10 + 3 \times \sqrt{10 - ....... \infty} } } \bigg)} \\\\\\ \sf \implies x = \sqrt{10 + 3 \sqrt{10 - 3x} }$

Squaring on both sides

$\\ \sf \implies x ^2 = 10 + 3 \sqrt{10 - 3x} \\\\\\ \sf \implies x^2 - 10 = 3 \sqrt{10 -3x}$

Squaring on both sides

$\\ \sf \implies (x^2 - 10)^2 = 9(10 -3x) \\\\\\ \sf \implies x^4 + 100 - 20x^2 = 90 - 27x \\\\\\ \sf \implies x^4 - 20x^2 + 27x + 100 - 90 = 0 \\\\\\ \sf \implies x^4 - 20x^2 + 27x + 10 = 0 \\\\\\ \sf \implies (x - 2)(x^3 + 2x^2 - 16x - 5) = 0 \\\\\\ \sf \implies (x - 2)(x + 5)(x^2 - 3x - 1) = 0 \\\\\\ \sf \implies x - 2 = 0 \: or \: x + 5 = 0 \: or \: x^2 - 3x - 1 = 0 \\\\\\ \sf \implies x = 2 \: or \: x = - 5 \: or \: x^2 - 3x - 1 =0$

(Neglect 2 and - 5 since 2 is not there in options and x is under root so x can't be negative )

$\\ \sf \implies x^2 - 3x - 1 = 0$

Using quadratic formula i.e, {-b ± √(b²-4ac)}/2a

and here a = 1, b = - 3, c = - 1

$\sf \implies x = \dfrac{ - ( - 3) \pm \sqrt{ ( - 3)^2 - 4(1)( - 1)} }{2(1)} \\\\\\ \sf \implies x = \dfrac{3 \pm \sqrt{9 + 4} }{2} \\\\\\ \sf \implies x = \dfrac{3 \pm \sqrt{13} }{2} \\\\\\ \sf \implies x = \dfrac{3 + \sqrt{13} }{2} \: or \: x = \dfrac{3 - \sqrt{13}}{2}$

(Neglecting x = (3 - √13)/2 )

$\sf \implies x = \dfrac{ \sqrt{13} + 3}{2}$

Therefore the answer is c) (√13 + 3)/2.