Text description - screencast for exercise 5

[The narrator reads out the onscreen text.]

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.

[The equation expands and simplifies to the discriminant equals 4 times (k plus 1) all squared, minus 4 times (4k - 8k squared plus 1 minus 2k). So the discriminant equals 4 times (k squared plus 2k plus 1) minus 4 times (negative 8 k squared plus 2k plus 1. So the discriminant equals k squared plus 8k plus 4 plus 32k squared minus 8k minus 4. So the discriminant equals 36k squared.]

NARRATOR: 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.

[On an x and y graph, there is a curved line below the x-axis. Its left-hand side crosses the y-axis.]

NARRATOR: 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.