Question

Resolva as seguintes equações?

1) x²-3√3x+6=0

2) x+1/x=3

3)(-x²+1).(x²-3x+2)=0

Answer 1

1)$x^{2} -3. \sqrt{3} x+6=0$
   $x^{2} - \sqrt{27} x+6=0$
   Δ = (√27)² - 4.6
   Δ = 27 - 24
   Δ = 3

$x= [-(-3 \sqrt{3} ) +- \sqrt{3} ]/ 2$
$x^{I} =[3 \sqrt{3} + \sqrt{3} ]/2 =[4 \sqrt{3}] /2 = 2 \sqrt{3}$
$x^{II} =[3 \sqrt{3} - \sqrt{3} ]/2 = [2 \sqrt{3} ]/2= \sqrt{3}$
$S=[2 \sqrt{3} , \sqrt{3} ]$

2)$x+1/x=3$
$x^{2} +1=3x$
Δ=(-3)² -4.1.1
Δ=9-4
Δ=5

$x=[-(-3)+- \sqrt{5} ]/2$
$x^{I} =[3+ \sqrt{5} ]/2$
$x^{II} = [3 - \sqrt{5} ]/2$
$S=[ [3+ \sqrt{5} ]/2,[ 3-\sqrt{5} ]/2]$

3)Ou -x²+1=0, ou x²-3x+2=0
$-x^{2} +1=0$
$- x^{2} =-1$
$-1^{2} =-1$
$-1=-1$
$x^{I} = 1$

$x^{2} -3x+2=0$
$1+2=3(b)$
$1.2=2(c)$
$x^{II} =2$

$S=[1,2]$