Question

Find the value of k of which have real and equal roots:

1) X^2 - 2(k+1)x + k^2 = 0

Answer 1

Value of k= -1/2
To know how it is obtained ,refer to the attachment

Answer 2

Hey there !!!!!!!!

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A quadratic equation of the form Ax²+Bx+C=0 has

✪ Imaginary roots  if discriminant  < 0

                          B²-4AC < 0

✪ Real and equal roots if discriminant = 0

                          B²-4AC=0

✪ Real and distinct roos if discriminant > 0

                          B²-4AC >0

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x²-2(k+1)x+k² comparing with Ax²+Bx+C = 0

   A=1    B=-2(k+1)     C=k²

       B²-4AC=0

      (-2(k+1)²)-4*k²=0

       4(k+1)²-4k²=0

        4((k+1)²-k²)=0

       ((k+1)²-k²)=0

      k²+2k+1-k²=0

         2k+1=0

          k= -1/2

At k=-1/2   p(x)=  x²-2(k+1)x+k²  has real and equal roots.

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Hope this helped you.............