The quadratic case in practice

Hello. Welcome back to a step in practice. We are dealing here with the quadratic case. In exercise one, we consider the following inequality– 3x minus 2 over 2, strictly less than x square minus 2. Well, if we multiply by 2, this is equivalent to 2 x square minus 4 greater than 3x minus 2, which is 2 x square minus 3x minus 2 strictly greater than 0. Now, the discriminant of this polynomial of second degree is 3 to the square minus 4 times 2 times minus 2, which is 3 to the square plus 16. So we get 5 to the square, 25. Well, there are therefore two roots of the polynomial.

And the roots are 3 plus or minus 5 over 4. And so we get two roots.

When we get the plus, we obtain 2. With the minus, we get minus 1/2.

Now, what do we know about second degree inequalities? Well, here the sign of the coefficient of x square is strictly positive. And therefore, the polynomial is strictly positive everywhere except between the two roots. So the inequality is satisfied out of the interval between the two roots, and thus the solution is the union of the two intervals from minus infinity to 1/2 – minus 1/2, union 2 plus infinity. We do not take 2 or minus 1/2 because the polynomial vanishes at those points. In exercise two, we are studying a strict second degree inequality. More precisely, minus x square plus 6x minus 9, strictly positive. Now, the sign of the coefficient of x square is strictly negative.

Therefore, the polynomial of second degree is strictly negative everywhere, except between its roots, if it has roots. Now let us look at the discriminant of this polynomial. The discriminant is 6 to the square minus 4 times minus 1 times minus 9. That is 36, minus 36, 0. Therefore, there is just one root. And the root is minus 6 divided by minus 2. So 6 over 2, 3.

Therefore, the polynomial is strictly negative everywhere, except in 3, where it vanishes. Therefore, the inequality is never satisfied. So minus x square plus 6x minus 9 is strictly negative for all x in R minus 3, and is 0 for x equal to 3.

Therefore, the inequality does not have a solution. It’s the empty set. In exercise three, we are looking for the values of a parameter k in such a way that the inequality x square minus kx minus 20 strictly negative has at least a solution. Now, if we look at this polynomial, the sign of the coefficient of x square is strictly positive. Therefore, the polynomial is strictly positive everywhere, except at most between the roots, if it has roots. So this inequality is satisfied whenever this polynomial has two distinct roots. So if it has a strictly positive discriminant. So this has a solution if and only if the discriminant is strictly positive.

Well, how about the discriminant, the discriminant is k to the square minus 4 times minus 20, which gives k to the square plus 80, which is a strictly positive real number, for any k. Thus, the inequality has a solution for every value of k, for every k.

And this concludes these step in practice. Exercise four is not solved here, but you’ll find the solution in the PDF file. See you in the next step.