Question

resolva as equações irracionais:

a)raiz quadrada de 3x+6=x+2 ;

b) raiz quadrada de 4x+1= 2x-1

Answer 1

$\sqrt{3x+6} =x+2$

$( \sqrt{3x+6} )²=(x+2)²$

3x+2=x²+4x+4
x²+x+2=0
Δ=1+8=9
√Δ=3

$X'= \frac{-1+3}{2} = \frac{2}{2} = 1$

$x"= \frac{-1-3}{2} = \frac{-4}{2} = -2$

verificação
p/x=1

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

$\sqrt{9} =3$

3=3  V

p/ x=-2

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

$\sqrt{0} = 0$   V

S={-2,1}



b) $\sqrt{4x+1} = 2x-1$

$( \sqrt{4x+1} )² =( 2x-1$)²

4x+1 =4x² -4x+1

4x² -8x=0
4x(x-2)=0
4x=0
x=0

x-2=0
x=2

verf/

p/ x=0

$\sqrt{4(0)+1} =2(0) -1$

1≠-1

p/ x=2

$\sqrt{4(2)+1} =2(2)+1$

$\sqrt{9} =3$

3=3  V

S={2}