Show that the roots of this quadratic equation are real and distinct for all the real values of k.
This was how I did it.
a = (K + 5), b = 13, c = -6k
Using b² - 4ac,
Using the formula of finding the roots of a quadratic equation,
After this, I don't know what to do. I know that this way is wrong so please correct me on the steps.
I'm not sure that the bolded word should be there.
If the roots of are real and distinct, then it should be , not .This was how I did it.
a = (K + 5), b = 13, c = -6k
Using b² - 4ac,
Using the formula of finding the roots of a quadratic equation,
When solving a quadratic inequality, the first thing I do is change it to an equation anyway to find the critical points. So assuming that you plugged in numbers into the quadratic formula right:
This tells you that the quadratic has no real roots, that it doesn't cross the x-axis. That means that for any value of k, the expression will always be positive or always be negative. Since the leading coefficient is positive, for any k. Therefore, the roots of are real and distinct.
01
The discriminant of the quadratic equation
,
as you have shown, is
.
Recall that for the discriminant b² - 4ac:
if b² - 4ac > 0, there are two distinct real roots,
if b² - 4ac = 0, there is one real double root, and
if b² - 4ac < 0, there are no real roots.
Since we have shown that for any real number k, there will be two distinct real roots in the quadratic , for any real number k.
01