Given x^2-7x+12=0 What is the value of k such that the equation has 2 imaginary solutions?

Question

Parabola formula: y-k=a(x-h)^2

Answer 1

You have two different questions here:

The equation x^2 - 7x + 12 = 0 is able to be factored into (x-3)(x-4).

[ I factored the equation by looking at the second sign which is a +. This means that I need two numbers which add to 7 and have a product of 12. These two numbers are 3 and 4. The first sign being a - means tht both of these numbers are -. ]

This factoring means x = 3 or 4 for this equation.

The second question is understanding a parabola equation of the form (y-h) = a( x-k)^2. The vertex of the parabola is (k,h). If "a" is positive then the parabola has a U shape with both ends going up and if the a is negative then the parabola has a U shape with both ends going down. To have two different answers a parabola must not be symmetrical around the y -axis thus k cannot = 0. To have two different imaginary answers the parabola legs must not intersect the x axis. For positive "a" this means that k> 0 and for "a" negative a then k< 0.