Question

Resolva as equações polinomiais do 2° grau:
a)9x² - 3 = 0
b)10x² - 1000 = 0
c)5x² - 125 = 0
d)-3x² + 27 = 0

ajuda pfvr​

Answer 1

Resposta:

Explicação passo a passo:

$a)$

 $9x^{2} -3=0$

 $9x^{2} =3$

   $x^{2} =\frac{3}{9}$

   $x=\pm\:\sqrt{\frac{3}{9} }$

  $x=\frac{\sqrt{3} }{\sqrt{9} }$

  $x=\pm\:\frac{\sqrt{3} }{3}$

$b)$

 $10x^{2} -1000=0$

 $10x^{2} =1000$

    $x^{2} =1000/10$

    $x^{2} =100$

    $x=\pm\:\sqrt{100}$

    $x=\pm\:10$

$c)$

  $5x^{2} -125=0$

  $5x^{2} =125$

    $x^{2} =125/5$

    $x^{2} =25$

    $x=\pm\:\sqrt{25}$

    $x=\pm\:5$

$d)$

  $-3x^{2} +27=0$

     $3x^{2} =27$

       $x^{2} =27/3$

       $x^{2} =9$

       $x=\pm\:\sqrt{9}$

       $x=\pm\:3$

Answer 2

Resposta:

a) $9x^2 - 3 = 0$

$9x^2 = 3\\x^2=\dfrac{3}{9}\\x=\pm \sqrt{\dfrac{3}{9}}\\x_1=\dfrac{\sqrt{3}}{3};\ x_2=-\dfrac{\sqrt{3}}{3}$

b) $10x^2 - 1000 = 0$

$10x^2=1000\\x^2=\dfrac{1000}{10}\\x=\pm \sqrt{\dfrac{1000}{10}}\\x=\pm \sqrt{100}\\\\x_1=10;\ x_2=-10$

c) $5x^2 - 125 = 0$

$5x^2 =125\\x^2=\dfrac{125}{5}\\x^2=25\\x=\pm \sqrt{25}\\x_1=5;\ x_2=-5$

d) $-3x^2 + 27 = 0$

$-3x^2 =-27\\x^2=\dfrac{-27}{-3}\\x^2=9\\x=\pm \sqrt9\\x_1=3;\ x_2=-3$