Let-n-be-a-positive-integer-such-that-one-of-the-roofs-of-the-quadratic-equation-4x-2-4-3-4-x-3-n-24-0-is-an-integer-Find-the-value-of-n-

Question Number 8957 by Sopheak last updated on 07/Nov/16

L e t n b e a p o s i t i v e i n t e g e r s u c h t h a t o n e o f

t h e r o o f s o f t h e q u a d r a t i c e q u a t i o n

4 x^{2} - (4 \sqrt{3} + 4) x + \sqrt{3} n - 24 = 0 i s a n i n t e g e r

F i n d t h e v a l u e o f n

Commented by prakash jain last updated on 07/Nov/16

\frac{(4 \sqrt{3} + 4) \pm \sqrt{{(4 \sqrt{3} + 4)}^{2} - 4 \cdot 4 (\sqrt{3} n - 24)}}{8}

= \frac{4 (1 + \sqrt{3}) \pm 4 \sqrt{4 + 2 \sqrt{3} - \sqrt{3} n + 24}}{8}

= \frac{(1 + \sqrt{3}) \pm \sqrt{28 + (2 - n) \sqrt{3}}}{2}

if one of the roots is an integer then

\sqrt{28 + (2 - n) \sqrt{3}} contain \pm \sqrt{3} to

cancel out \sqrt{3} term in (1 + \sqrt{3})

\sqrt{28 + (2 - n) \sqrt{3}} = a \pm \sqrt{3}

2 a = \pm (2 - n) (i)

a^{2} + 3 = 28 \Rightarrow a = \pm 5 (ii)

from (i)

a = 5 \Rightarrow n = 12 o r n = - 8

a = - 5 \Rightarrow n = 12 o r n = - 8

n = - 8 is invalid since n > 0 .

h e n c e n = 12

c h e c k

4 x^{2} - 4 (1 + \sqrt{3}) x + 12 (\sqrt{3} - 2) = 0

x^{2} - (1 + \sqrt{3}) x + 3 (\sqrt{3} - 2) = 0

x^{2} - 3 x + (2 - \sqrt{3}) x - 3 (2 - \sqrt{3}) = 0

(x - 3) (x + 2 - \sqrt{3}) = 0

x = 3 is an integer solution .