Question

Find all real and complex solutionsa)x²+2x+1=0b)x²+4=0c)9x²-4x-14=0d)8x²+4x+32=0

Answer 1

Step-by-step explanation:

a). x²+2x+1=0

   $\left ( x + 1 \right )^{2}=0$

   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 =  $\frac{-4\pm \sqrt{-4^{2}-(4\times 9\times -14)}}{2\times 9}$

   = $\frac{-4\pm \sqrt{16+504}}{18}$

   = $\frac{-4\pm \sqrt{520}}{18}$

   = $\frac{-4\pm 22.8}{18}$

 x = $\frac{-4- 22.8}{18}$

    = -1.48

x = $\frac{-4+ 22.8}{18}$

  = 1.04

Therefore, x =  1.04 ,  -1.48

Hence the roots are real

d) 8x²+4x+32=0

x =  $\frac{-4\pm \sqrt{4^{2}-(4\times 8\times 32)}}{2\times 8}$

   = $\frac{-4\pm \sqrt{16-1024}}{16}$

   = $\frac{-4\pm \sqrt{1008}}{16}$

   = $\frac{-4\pm 31.7}{16}$

 x = $\frac{-4- 31.7}{16}$

    = -2.2

x = $\frac{-4+ 31.7}{16}$

  = 1.73

Therefore, x =  1.73 ,  -2.2

Hence the roots are real