Question-69018

Question Number 69018 by Maclaurin Stickker last updated on 17/Sep/19

Commented by Prithwish sen last updated on 18/Sep/19

It is only possible when

4 a + 5 b + 1 = 8 ab + 1

\Rightarrow 4 a + 5 b = 8 ab \dots \dots \dots \dots (i)

and

(16 a^{2} + b^{2} + 1) (4 a + 5 b + 1) = {(4 a + 5 b + 1)}^{2} = ({(8 ab + 1)}^{2} \dots \dots (ii)

\Rightarrow {16 a}^{2} + b^{2} + 1 = 4 a + 5 b + 1

{16 a}^{2} + b^{2} + 1 = 8 ab + 1 from (i)

{(4 a - b)}^{2} = 0

\Rightarrow 4 a = b \dots \dots . (ii)

∴ from (i) and (ii) we get

a = 0, b = 0 and a = \frac{3}{4} and b = 3

∵ the equation only satified by

a = \frac{3}{4} and b = 3

∴ the possible solution is (\frac{3}{4}, 3)