Respuesta :
Step-by-step explanation:
a). x²+2x+1=0
[tex]\left ( x + 1 \right )^{2}=0[/tex]
x = -1 , - 1
Therefore the roots are real.
b). x²+4=0
x²= -4
x = 2 , -2
Therefore the roots are real.
c) 9x²-4x-14 =0
x = [tex]\frac{-4\pm \sqrt{-4^{2}-(4\times 9\times -14)}}{2\times 9}[/tex]
= [tex]\frac{-4\pm \sqrt{16+504}}{18}[/tex]
= [tex]\frac{-4\pm \sqrt{520}}{18}[/tex]
= [tex]\frac{-4\pm 22.8}{18}[/tex]
x = [tex]\frac{-4- 22.8}{18}[/tex]
= -1.48
x = [tex]\frac{-4+ 22.8}{18}[/tex]
= 1.04
Therefore, x = 1.04 , -1.48
Hence the roots are real
d) 8x²+4x+32=0
x = [tex]\frac{-4\pm \sqrt{4^{2}-(4\times 8\times 32)}}{2\times 8}[/tex]
= [tex]\frac{-4\pm \sqrt{16-1024}}{16}[/tex]
= [tex]\frac{-4\pm \sqrt{1008}}{16}[/tex]
= [tex]\frac{-4\pm 31.7}{16}[/tex]
x = [tex]\frac{-4- 31.7}{16}[/tex]
= -2.2
x = [tex]\frac{-4+ 31.7}{16}[/tex]
= 1.73
Therefore, x = 1.73 , -2.2
Hence the roots are real
