WEBVTT
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[MUSIC PLAYING]
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Hello.
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Let us solve the first exercise
of the equivalence in practise
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step.
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We want to solve this equation.
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OK, first of all,
we have to compute
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the domain of this equation.
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It's clear that here we have a
problem with this denominator
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and also with this denominator.
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And we have to choose
as domain the biggest
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subset of real numbers where
these two denominators do not
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vanish.
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The domain is clearly equal to
the set of real numbers minus 0
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and minus 1.
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OK.
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Inside this domain, this
equation always makes sense.
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Now, remember that starting with
one equation, if we multiply
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on both sides this equation
by something which is never
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0 inside the domain, then
we get an equation which is
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equivalent to the starting one.
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What means to be equivalent?
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It means that the two equations
have the same set of solutions
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inside the domain.
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In such a case, it's a good
idea to multiply on both sides
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by x times x plus 1.
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What do we get?
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You see, if I multiply this
term by x times x plus 1,
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I get just x plus 1.
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And if I multiply this
term by x times x plus 1,
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I get x times x plus 3/2.
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And now if I multiply 0 by x
times x plus 1, I get 0, OK?
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Let us do some just
easy computations.
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And here we get x squared plus
1 plus 3/2 x plus 1 equals 0.
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That is, x squared plus
5/2 x plus 1 equals 0.
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And now we have a
new equation which
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is equivalent to the starting
one inside the domain.
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OK, let us compute the solutions
of this equation of degree 2.
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First of all, let us
compute the discriminant
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of this polynomial.
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And we get the
square of 5/2, which
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is 25/4, minus 4 times the
leading coefficient, which
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is 1, times the constant
term, which is again 1.
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And we get, taking 4 as common
denominator, 25 minus 16,
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which is 9/4.
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Therefore, the solutions
of this equation
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are x equal to minus
5/2 plus or minus
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the square root of the
discriminant over 2
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times the leading
coefficient, which is 1.
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That is, we get minus 5/2
plus or minus 3/2 over 2,
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and this is minus 5/2
plus 3/2 is minus 2/2,
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which is minus 1/2,
which is minus 1/2.
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And the minus 5/2 minus 3/2 is
minus 8/2, which is minus 4/2
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is minus 2.
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OK.
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Now look, both
these two solutions
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belong to the domain of
our initial equation.
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Therefore, because this
equation and this equation
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are equivalent
inside this domain,
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we can conclude that
our initial equation
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has minus 1/2 and
minus 2 as solutions.
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Solutions are minus
1/2 and minus 2.
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Thank you for the attention.
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