What is the value of k? Does the unary quadratic equation x 2+2 (k+1) x+2k 2-k+3 = 0 have two equal real roots? The unary quadratic equation (k-2) x 2+3x+ 1 = 0 has a real root, and the solution of the value range of k is 1. The equation has two equal real roots, namely the discriminant δ= 0.

From the meaning of the question, Delta = 4 (k+1) 2-4 (2k 2-k+3) = 0, so we can get k= 1 or k=2.

When k= 1, the equation is x 2+4x+4 = 0, and both equal real roots are x =-2;

When k=2, the equation is x 2+6x+9 = 0, and both equal real roots are x=-3.

2. First of all, the equation is a quadratic equation, so the quadratic coefficient k-2 is not 0, that is, k is not equal to 2.

Secondly, if the equation has real roots, then the discriminant δ >; =0. Therefore

Discriminant δ= 9-4(k-2)>=0, k

The value range of k is K.