No real solution or No solution?

I'm confused about what y=x means in your question. It looks like you are trying to solve the equation: f(x) = x^2 +kx +1 = 0 This equation will ALWAYS have two solutions, either both real, or both complex. (Duplicate solutions count as two, like for 0 = (x-1)^2, whichi is of the same form.) Ignore the case where k is complex (better term than "imaginary").

You're overanalyzing the question. You are most likely working in the area of real polynomial functions. Besides, if k is complex, then the equation has two complex solutions.

The solutions are -k +/- sqrt(k^2 - 4) / 2. The discriminant is D = k^2 -4, not the expression you gave. Assuming k is real: If D is 0, then |k| =2 and there's a double, real root.

If D is positive, then |k| >2 and the square root part of the solutions is real, and there are two distinct real roots If D is negative, then |k| 0, it moves down and to the left. When |k| 2, part of the graph lies below the x-axis, and the equation has two distinct real solutions. Comments after your additional explanation: You still have the discriminate wrong.

It's b^2 -4ac a = 1 be = (k-1) c = 1 D = (k-1)^2 -4. You could leave it like that, or multiply out k^2 -2k -3 (not +5) Here's a new picture, of f(x) with k as before, and the line y = x (note Excel doesn't draw this to scale.) i276.photobucket.com/albums/kk2/f… The dotted lines, for k = -1 or 3, the graphs are tangent to the line, so there is one point of intersection. These correspond to a double real solution to the modified equation.

The very heavy lines cut the line x=2 in two points, corresponding to two real roots of the modified equation. The other graphs miss the line x=y, and correspond to complex solutions to the equation, and therefore no intesection in the xy-plane. The modification of the equations just changes the existence of solutions from |k| >= 2 to |k-1| >=2.

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