WEBVTT
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Ciao.
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Let us solve the first exercise
of the polynomial equations
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in practice step.
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We have to solve this
polynomial equation.
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It is a polynomial equation,
depending on the parameter,
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the parameter k.
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Let us see how to
solve this problem.
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First of all, let
us try to compute
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the degree of this equation.
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You immediately realize
that if k is equal to 1,
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then this term is 0.
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And therefore the
term of degree 2
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disappears, and we'll remain
with an equation of degree 1.
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Precisely, we remain with the
equation x plus 1 equals 0,
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and this equation, as all
the equations of degree 1,
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has exactly one solution,
precisely, in this case,
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x equal to minus 1.
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On the other case, if k
is not equal to 1, then
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here we have an
equation of degree 2.
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To solve this
equation, as usual,
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we start computing
the discriminant
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of this polynomial.
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And we have that the
discriminant is the square of k
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minus 4 times the
leading coefficient
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times the constant
term, which is 1.
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What do we get?
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We get k squared minus
4 times k plus 4.
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You immediately
realize that this
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is exactly equal to
k minus 2 square,
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and therefore, it is always
greater or equal than 0.
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Therefore, our
equation will have
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always solutions, but how many?
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If delta is equal to 0--
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the discriminant is equal to 0--
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then we have exactly
one solution.
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And this happens
when k is equal to 2.
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If delta is greater
than 0, then we
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have two solutions,
two distinct solutions.
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And this happens if and
only if k is not equal to 2.
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Let us compute the solutions
in these two cases.
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First case, then, k equals 2.
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We know that we have
just one solution,
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and we chase this solution.
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x equals-- it is minus
k, but k is equal to 2.
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Therefore it's minus 2, plus
or minus the square root
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of the discriminant.
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But the discriminant is
0, and therefore this
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will not give any contribution.
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Over 2 times the
leading coefficient.
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But k is equal to 2.
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Therefore the leading
coefficient is 2 minus 1,
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that is, 1--
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2 times 1.
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And then we get that the
only solution is minus 2
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over 2, which is minus 1.
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Ok, and what happens
if k is not equal to 2?
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Then in this case, we have
that discriminant is positive,
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and we have two solutions.
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Let us compute these solutions.
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We have that x is equal to what?
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And now you see the parameter
k is not determined univocally.
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And then we have that
the solutions are minus
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k plus or minus the square
root of the discriminant, which
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is k minus 2 squared, over 2
times the leading coefficient.
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And this is equal to what?
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Attention now.
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The square root of
k minus 2 square
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is the module of k minus 2.
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But we have a plus or minus
in front of this module,
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and therefore we can
forget about the module.
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And we get minus
k plus or minus k
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minus 2 over 2 times k minus 1.
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And this is equal to--
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ok, we have minus k
plus or minus k minus 2
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over 2k minus 2.
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Which is, ok, if I
choose the plus here,
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I have minus k plus k minus
2 over 2 times k minus 2.
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Therefore, it remains 2 over
2k minus 2, which is equal,
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dividing by 2, both the
numerator and denominator,
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we get 1 over k minus 1.
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And if we choose
minus, here we have
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minus k plus 2, which is minus 2
times k plus 2 over 2k minus 2.
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And you can easily realize
that this numerator
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and this denominator are one
the opposite of the other,
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and therefore we remain
exactly with minus 1.
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Then we have that if
k is not equal to 2,
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always in the case
k not equal to 1--
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that is, in the case in which we
have an equation of degree 2--
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then we have these
two solutions.
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Then we can summarize
the situation,
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and we can conclude, saying
that if k is equal to 1,
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then we have one solution,
x equal to minus 1.
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If k is equal to 2, then
we have one solution,
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x equal to minus 1 again.
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And in all the other cases,
k different from 1 and 2,
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we have two distinct solutions,
and they are one x equal
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to 1 over k minus 1, this
solution depending on k.
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And the other is x
equal to minus 1.
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Thank you for your attention.
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