Find k, so that the quadratic equation (k+1)x 2 −2(k+1)x+1=0 has equal roots.

Answer 1

set the discriminant = 0 and solve for k

b^2-4ac=0

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

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

k^2 +2k +1 = k+1

k^2 +2k =k

k^2 +k = 0

k(k+1) = 0

k = 0 or -1

but if k=-1, it's no longer a quadratic equation, so k=0 is the only real solution.

if you used k=-1 as a solution the "equation" collapses into a contradiction 1=0 which is no longer quadratic or an equation.

x^2 -2x +1 = 0

(x-1)^2 = 0

set each factor = 0

x-1=0

there is one root, zero or x intercept with multiplicity 2 or call it

"two roots" that repeat or are equal to each other

x=1, 1

or just

x=1

graphically the quadratic equation is an upward opening parabola with vertex (1,0)

the parabola is tangent to the x axis at its vertex which is also the minimum point on the parabola