Question

solve this problems (algebra, polynomial)

Answer 1

Answer:

$\sqrt{x + \sqrt{4x \sqrt{16x + 3} } } = \sqrt{x - 1}$

squaring on both sides,

$x + \sqrt{4x \sqrt{16x + 3} } = x - 1$

$\sqrt{4x \sqrt{16x + 3} } = - 1$

again, squaring on both sides,

$4x \sqrt{16x + 3} = 1$

$\sqrt{16x + 3} = \frac{1}{4x}$

squaring on both sides,

$16x + 3 = \frac{1}{16 {x}^{2} }$

$3 = \frac{1}{16 {x}^{2} } - 16x$

$3 = \frac{1 - 16 {x}^{2} \times 16x }{16 {x}^{2} }$

$48 {x}^{2} = 1 - 256 { x}^{3}$

$256 {x}^{3} + 48 {x}^{2} = 1$