Polynomial equations in practice

Ciao. Let us solve the first exercise of the polynomial equations in practice step. We have to solve this polynomial equation. It is a polynomial equation, depending on the parameter, the parameter k. Let us see how to solve this problem. First of all, let us try to compute the degree of this equation. You immediately realize that if k is equal to 1, then this term is 0. And therefore the term of degree 2 disappears, and we’ll remain with an equation of degree 1. Precisely, we remain with the equation x plus 1 equals 0, and this equation, as all the equations of degree 1, has exactly one solution, precisely, in this case, x equal to minus 1.

On the other case, if k is not equal to 1, then here we have an equation of degree 2. To solve this equation, as usual, we start computing the discriminant of this polynomial. And we have that the discriminant is the square of k minus 4 times the leading coefficient times the constant term, which is 1. What do we get? We get k squared minus 4 times k plus 4. You immediately realize that this is exactly equal to k minus 2 square, and therefore, it is always greater or equal than 0. Therefore, our equation will have always solutions, but how many? If delta is equal to 0– the discriminant is equal to 0– then we have exactly one solution.

And this happens when k is equal to 2.

If delta is greater than 0, then we have two solutions, two distinct solutions. And this happens if and only if k is not equal to 2. Let us compute the solutions in these two cases. First case, then, k equals 2. We know that we have just one solution, and we chase this solution. x equals– it is minus k, but k is equal to 2. Therefore it’s minus 2, plus or minus the square root of the discriminant. But the discriminant is 0, and therefore this will not give any contribution. Over 2 times the leading coefficient. But k is equal to 2. Therefore the leading coefficient is 2 minus 1, that is, 1– 2 times 1.

And then we get that the only solution is minus 2 over 2, which is minus 1.

Ok, and what happens if k is not equal to 2? Then in this case, we have that discriminant is positive, and we have two solutions. Let us compute these solutions. We have that x is equal to what? And now you see the parameter k is not determined univocally. And then we have that the solutions are minus k plus or minus the square root of the discriminant, which is k minus 2 squared, over 2 times the leading coefficient.

And this is equal to what? Attention now. The square root of k minus 2 square is the module of k minus 2. But we have a plus or minus in front of this module, and therefore we can forget about the module. And we get minus k plus or minus k minus 2 over 2 times k minus 1. And this is equal to– ok, we have minus k plus or minus k minus 2 over 2k minus 2.

Which is, ok, if I choose the plus here, I have minus k plus k minus 2 over 2 times k minus 2. Therefore, it remains 2 over 2k minus 2, which is equal, dividing by 2, both the numerator and denominator, we get 1 over k minus 1. And if we choose minus, here we have minus k plus 2, which is minus 2 times k plus 2 over 2k minus 2. And you can easily realize that this numerator and this denominator are one the opposite of the other, and therefore we remain exactly with minus 1.

Then we have that if k is not equal to 2, always in the case k not equal to 1– that is, in the case in which we have an equation of degree 2– then we have these two solutions. Then we can summarize the situation, and we can conclude, saying that if k is equal to 1, then we have one solution, x equal to minus 1. If k is equal to 2, then we have one solution, x equal to minus 1 again. And in all the other cases, k different from 1 and 2, we have two distinct solutions, and they are one x equal to 1 over k minus 1, this solution depending on k.

And the other is x equal to minus 1. Thank you for your attention.