Question

Resolva as equaçoes irracionais abaixo:

A)√x+1=2
B)x-6=√x
C)√x+5=x-1
D)√3x+16 -x=2
E)2+1√2x-1=x
F)X-1=√4x+5

Answer 1

a)

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

b)

$x - 6 = \sqrt{x} \\ {(x - 6)}^{2} = { (\sqrt{x}) }^{2} \\ {x}^{2} - 12x + 36 = x \\ {x}^{2} - 13x + 36 = 0 \\ (x - 4)(x - 9) = 0 \\ x = 9 \\ x = 4$

c)

$\sqrt{(x + 5)} = x - 1 \\ { (\sqrt{x + 5}) }^{2} = {(x - 1)}^{2} \\ x + 5 = {x}^{2} - 2x + 1 \\ {x}^{2} - 3x - 4 = 0 \\ (x + 1)(x - 4) = 0 \\ x = 4 \\ x = - 1$

d)

$\sqrt{3x + 16} - x = 2 \\ \sqrt{3x + 16} = x + 2 \\ {( \sqrt{3x + 16} )}^{2} = {(x + 2)}^{2} \\ 3x + 16 = {x}^{2} + 4x + 4 \\ {x}^{2} + x - 12 = 0 \\ (x - 3)(x + 4) = 0 \\ x = 3 \\ x = 4$

e) não entendi

f)

$x - 1 = \sqrt{4x + 5} \\ {(x - 1)}^{2} = {( \sqrt{4x + 5} )}^{2} \\ {x}^{2} - 2x + 1 = 4x + 5 \\ {x}^{2} - 6x - 4 = 0 \\ x = \frac{6 + - \sqrt{36 - 4 \times 1 \times ( - 4)} }{2} \\ x = \frac{6 + - \sqrt{36 + 16} }{2} \\ x = \frac{6 + - 2 \sqrt{13} }{2} \\ x = 3 + - \sqrt{13} \\ x = 3 + \sqrt{13} \\ x = 3 - \sqrt{13}$