Question

me ajudem nessas questões de 2°grau... 30 pontos

A) 9ײ + 6×+1=0
B) 3ײ - 6×+9=0
c) 3× ² - +4×+9=0
d) ײ-10× +1=0
E) ײ-5×+6=0

Answer 1

a) $9x^2+6x+1=0$

$\Delta=6^2-4\cdot9\cdot1=36-36=0$

$x=\dfrac{-6\pm\sqrt{0}}{2\cdot9}=\dfrac{-6}{18}=-\dfrac{1}{3}$

$S=\left{-\dfrac{1}{3}\right}$

b) $3x^2-6x-9=0$

$\Delta=(-6)^2-4\cdot3\cdot(-9)=36+108=144$

$x=\dfrac{-(-6)\pm\sqrt{144}}{2\cdot3}=\dfrac{6\pm12}{6}=1\pm2$

$x'=1+2=3$

$x"=1-2=-1$

$S=\{-1,3\}$

c) $3x^2-4x+9$

$\Delta=(-4)^2-4\cdot3\cdot9=16-108=-92$

Não há soluções reais

d) $x^2-10x+1=0$

$\Delta=(-10)^2-4\cdot1\cdot1 =100-4=96$

$x=\dfrac{-(-10)\pm\sqrt{96}}{2\cdot1}=\dfrac{10\pm4\sqrt{6}}{2}=5\pm2\sqrt{6}$

$x'=5+2\sqrt{6}$

$x"=5-2\sqrt{6}$

e) $x^2-5+6=0$

$\Delta=(-5)^2-4\cdot1\cdot6=25-24=1$

$x=\dfrac{-(-5)\pm\sqrt{1}}{2\cdot1}=\dfrac{5\pm1}{2}$

$x'=\dfrac{5+1}{2}=3$

$x"=\dfrac{5-1}{2}=2$