WEBVTT
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[MUSIC PLAYING]
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Hello.
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Welcome back to a
step in practice.
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We are dealing here
with the quadratic case.
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In exercise one, we consider
the following inequality--
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3x minus 2 over 2, strictly
less than x square minus 2.
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Well, if we multiply
by 2, this is
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equivalent to 2 x square minus
4 greater than 3x minus 2,
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which is 2 x square
minus 3x minus 2
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strictly greater than 0.
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Now, the discriminant of this
polynomial of second degree
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is 3 to the square minus 4
times 2 times minus 2, which
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is 3 to the square plus 16.
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So we get 5 to the square, 25.
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Well, there are therefore
two roots of the polynomial.
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And the roots are 3
plus or minus 5 over 4.
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And so we get two roots.
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When we get the
plus, we obtain 2.
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With the minus,
we get minus 1/2.
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Now, what do we know about
second degree inequalities?
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Well, here the sign of the
coefficient of x square
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is strictly positive.
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And therefore, the polynomial
is strictly positive
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everywhere except
between the two roots.
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So the inequality is satisfied
out of the interval between
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the two roots, and thus
the solution is the union
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of the two intervals from
minus infinity to 1/2 --
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minus 1/2, union
2 plus infinity.
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We do not take 2 or minus 1/2
because the polynomial vanishes
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at those points.
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In exercise two, we are
studying a strict second degree
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inequality.
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More precisely, minus x
square plus 6x minus 9,
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strictly positive.
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Now, the sign of the
coefficient of x square
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is strictly negative.
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Therefore, the polynomial
of second degree
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is strictly negative everywhere,
except between its roots,
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if it has roots.
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Now let us look at
the discriminant
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of this polynomial.
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The discriminant is 6 to the
square minus 4 times minus 1
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times minus 9.
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That is 36, minus 36, 0.
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Therefore, there
is just one root.
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And the root is minus
6 divided by minus 2.
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So 6 over 2, 3.
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Therefore, the polynomial is
strictly negative everywhere,
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except in 3, where it vanishes.
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Therefore, the inequality
is never satisfied.
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So minus x square
plus 6x minus 9
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is strictly negative
for all x in R minus 3,
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and is 0 for x equal to 3.
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Therefore, the inequality
does not have a solution.
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It's the empty set.
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In exercise three,
we are looking
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for the values of a
parameter k in such a way
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that the inequality x
square minus kx minus 20
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strictly negative has
at least a solution.
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Now, if we look at
this polynomial,
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the sign of the coefficient of
x square is strictly positive.
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Therefore, the polynomial
is strictly positive
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everywhere, except at
most between the roots,
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if it has roots.
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So this inequality is satisfied
whenever this polynomial
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has two distinct roots.
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So if it has a strictly
positive discriminant.
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So this has a
solution if and only
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if the discriminant
is strictly positive.
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Well, how about
the discriminant,
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the discriminant is k to the
square minus 4 times minus 20,
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which gives k to
the square plus 80,
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which is a strictly positive
real number, for any k.
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Thus, the inequality has a
solution for every value of k,
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for every k.
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And this concludes
these step in practice.
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Exercise four is
not solved here,
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but you'll find the
solution in the PDF file.
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See you in the next step.
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