Question

Find the values of k for which the following equations have real and equal roots : x^2-2(k+1)x+k^2

Answer 1

equal roots means D =0
therefore b^2-4ac=0
=>(-2k-2)^2-4(1)(k^2)=0
=>4(k^2+1+2k)=4(k^2)
=>k^2+2k+1=k^2
=>2k=-1
=>k=-1/2

Answer 2

Given Equation is x^2 - 2(k + 1)x + k^2.

It is in the form of ax^2 + bx + c, we get

a = 1, b = -2(k + 1), c = k^2.

Given that the Equation has real and equal roots.

= > D = b^2 - 4ac = 0

         = (-2(k + 1)^2) - 4(1)(k^2) = 0

         = 4(k + 1)^2 - 4k^2 = 0

         = 4(k^2 + 1 + 2k) - 4k^2 = 0

         = 4k^2 + 4 + 8k - 4k^2 = 0

         = 4 + 8k = 0

         = 4 = 8k

        = k = -1/2.



Hope this helps!