Find the value(s) of k that make the roots real, rational, and equal. kx^2-14x+49=0
The roots are real when the part under the radical of the quadratic formula (called the determinant) is greater than or equal to zero, so \[b^2-4ac \ge 0\] a=k, b=-14, c=49 so we have \[(-14)^2-4k(49) \ge 0\]\[196-196k \ge 0\]\[k \le 1\]
not to be mean, but it's actually called a discriminant and i only know that because i have my notes...but i just wanted to tell you to make sure you knew for next or something...but is every equation always going to be set greater than or equal zero?
My bad, discriminant. Thanks for the correction, I don't use the term often. To answer your question, whenever we need to know if a root is real, we need to know if the number under the radical is negative or not. Hence we need find out if the discriminant is greater than or equal to zero. So yes we will always be asking\[b^2-4ac \ge 0\]in order to find the answer.
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