Question

RESOLVA AS EQUAÇÕES DO 2 GRAU,EM R:
COM OS CALCULOS:
x²+2x-3=0

2x²-5x+2=0

2x²-50=0

x²+9=0

3x²-8x=0

x²-x-6=0

Answer 1

$x^2+2x-3=0 \\ D=2^2-4.1.(-3) \\ D=4+12=16 \\ x= \frac{-2+- \sqrt16}{2.1} \\ x_1= \frac{-2+4}{2} = \frac{2}{2} =1 \\ x_2= \frac{-2-4}{2} = \frac{-6}{2} = -3 [tex]2x^2-5x+2=0 \\ D=(-5)^2-4.2.2 \\ D=25-16=9 \\ x= \frac{5+- \sqrt{9} }{2.2} \\ x_1= \frac{5+3}{4}= \frac{8}{4} = 2 \\ x_2= \frac{5-3}{4}=( \frac{2}{4})^{:2} = \frac{1}{2} = 0,5$

$2x^2-50=0 \\ D=0^2-4.2.(-50) \\ D=0+400 \\ x= \frac{0+- \sqrt{400} }{2.2} \\ x_1= \frac{20}{4}=5 \\ x_2= \frac{-20}{4}=-5$

$x^2+9=0 \\ D=0^2-4.1.9 \\ D=-36$ Não existe resposta para essa equação!

$x^2-x-6=0 \\ D=1^2-4.1.(-6) \\ D=1+24=25 \\ x= \frac{1+- \sqrt{25} }{2.1} \\ x_1= \frac{1+5}{2}= \frac{6}{2} = 3 \\ x_2= \frac{1-5}{2} = \frac{-4}{2} = -2$

$3x^2-8x=0 \\ D=(-8)^2-4.3.0 \\ D=64 \\ x =\frac{8+- \sqrt{64} }{2.3} \\ x_1= \frac{8+8}{6}= (\frac{16}{6})^{:2}= \frac{8}{3} \\ x_2 = \frac{8-8}{6}= \frac{0}{6}=0$