Question

raiz quadrada de (3x+1)+(raiz quadrada de (x-1)=6

Answer 1

$\sqrt{3x+1} + \sqrt{x-1} =6$

$( \sqrt{3x+1} + \sqrt{x-1})^{2} = 6^{2}$

$( \sqrt{3x+1})^{2} + 2.(\sqrt{3x+1}).(\sqrt{x-1})+( \sqrt{x-1})^{2} = 36$

$3x+1 + 2\sqrt{(3x+1).(x-1)}+x-1 = 36$

$4x+ 2\sqrt{(3x+1).(x-1)} = 36$

$2\sqrt{(3x+1).(x-1)} = 36-4x$

$\sqrt{(3x+1).(x-1)} = 18-2x$

$\sqrt{3 x^{2} -2x-1} = 18-2x$

$(\sqrt{3 x^{2} -2x-1} )^{2} = (18-2x)^{2}$

$3 x^{2} -2x-1 = 324-72x+4 x^{2}$

$x^{2} -70x+325 = 0$

$(x-5).(x-65)=0$

$x-5=0$ ⇒ $x=5$
ou
$x-65=0$ ⇒ $x=65$