Question

root over 4x+9 + root over 4x-9=5+root 7.........soln. for X

Answer 1

$\sqrt{4x + 9} + \sqrt{4x - 9} = 5 + \sqrt{7}$
squaring both sides,
$4x + 9 + 4x - 9 + 2 \sqrt{(4x + 9)(4x - 9)} = 25 + 7 + 10 \sqrt{7}$
$8x + 2 \sqrt{16 {x}^{2} - 81} = 32 + 10 \sqrt{7}$
$\sqrt{16 {x}^{2} - 81} = \frac{32 + 10 \sqrt{7} - 8x}{2}$
squaring both sides again,
$16 {x}^{2} - 81 = \frac{{(32 + 10 \sqrt{7} - 8x)}^{2}}{4}$
$right \: side = 625 + (10 \times 7) + 64 {x}^{2} + 12500 \sqrt{7} - (160 \sqrt{7} )x - 10000x$
$- 324 = 695 + 12500 \sqrt{7} - x(160 \sqrt{7} + 10000)$
$x(160 \sqrt{7} + 10000) = 695 + 324 + 12500 \sqrt{7}$

$x = \frac{1019 + 12500 \sqrt{7} }{160 \sqrt{7} + 10000}$
is my answer correct

Answer 2

Hope it's helpful........... ✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌