NARRATOR: Exercise five.
For what values of k does the equation
(4k + 1) times x squared minus 2(k + 1) times x + (1 – 2k) = 0
have one real solution?
For what values of k, if any, is the quadratic negative definite?
When considering the number of solutions of a quadratic equation
of the form ax squared + bx + c = 0,
we must consider the discriminant, b squared – 4ac.
In the case of this quadratic, a is equal to (4k + 1),
b is equal to –2(k + 1)
and c is equal to (1 – 2k).
So the discriminant can be calculated as
(-2(k + 1)) all squared, minus 4(4k + 1) times (1 – 2k).
And when we expand and simplify,
we find that the discriminant
is equal to 36k squared.
A quadratic has only one solution when its discriminant is 0.
In this case that is when 36k squared is equal to 0,
and this will happen when k = 0.
So therefore, this quadratic equation has one real solution
when k is equal to 0.
A quadratic is negative definite
if the graph of the quadratic sits entirely below the x-axis.
So this means that the y-coordinate of the vertex
would need to be negative
and the parabola would need to be inverted
so that the coefficient of x squared is negative.
And also we would note that the
quadratic would have no x-intercepts.
In the case of this quadratic, we know that the discriminant -
which tells us about the number of solutions of the equation
and hence the number of x-intercepts of the graph -
this discriminant is 36k squared,
which is always greater than or equal to 0,
no matter the value of k.
So if the discriminant cannot be negative,
this quadratic will always have x-intercepts
and hence it cannot be negative definite.