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brilliants [131]
3 years ago
11

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

Mathematics
1 answer:
Sati [7]3 years ago
4 0

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

 

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