Question

prove that the root of x^2+(1-k)x+k-3=0 are real for all real values of k​

Answer 1

The roots of a quadratic equation are real whenever the discriminant is non-negative. For the function given, the discriminant is

(1+k)2–4(1)(−k) , which simplifies as

k2+6k+1 .

To determine when this is non-negative, we note that its zeros are at

−6±36–4√2 , which are

−3–22–√ and −3+22–√ . Since this expression is quadratic, it can be graphed as a parabola which opens upward, and so is non-negative to the left of the first zero, and to the right of the second zero. Consequently, the roots of the original function are real for values of k in

(−∞,−3–22–√]∪[−3+22–√,∞) , so NOT for all values of k.