WEBVTT
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[MUSIC PLAYING]
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Hello.
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Let us consider now the
fourth exercise of the section
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introduction:
types of equations.
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We wanted to understand where
the graphs of two functions
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intersect.
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The two functions are f, which
sends x to 6 times x plus 2,
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and g, which sends x
to x squared plus 9.
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First of all, let us observe
that these two functions are
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defined on all the real numbers.
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Therefore their
domains are equal,
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and both are equal to the
set of all real numbers.
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Now, which are the
points of the graph of f?
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OK, a generic point of the graph
of f has the following shape--
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x, f of x, where x
is any real number.
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That is point of the
type x 6 times x plus 2.
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And the points of the graph of
g have the following shape--
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x, g of x, which is equal
to x, and x squared plus 9.
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What means to intersect
the two graphs?
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I am looking for points in the
plane of the following shapes--
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of both the following shapes.
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Therefore, I am looking
for real numbers--
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x-- such that 6 times x plus 2
is equal to x squared plus 9.
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That is, we get the
equation 6 times x plus 2
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equal to x squared plus 9.
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OK, and this equation
is equivalent to what?
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OK, we just rearrange
the terms, and we
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get x squared minus 6 times
x, and we get plus 7 equal 0.
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OK, this is an equation
of second degree.
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Let us compute its
discriminant, delta.
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Then it is equal to what?
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To the square of
the minus 6, which
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is 36, minus 4 times the product
of the leading coefficient,
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which is 1, by the
constant terms, which is 7.
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Then we get 36 minus
28, which is 8.
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Observe that the discriminant
is greater than 0.
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What it means?
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It means that this equation
has two different solutions.
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And in terms of our
exercise, what means?
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It means that we will
find two different points
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of intersection
between the two graphs.
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Let us solve the equation.
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We have that the solutions of
our equation are minus minus 6,
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which is 6, plus or minus the
square root of the discriminant
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divided by 2 times the
leading coefficient,
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and this is equal to what?
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OK, 6 divided by 2
is 3, plus or minus.
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And now we have the
square root of 8.
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The square root of 8 is
exactly like to write--
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the square root of 4 times 2.
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And we have to divide by 2.
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Now, this is equal to what?
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3 plus minus, and the
square root of 4 is 2.
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Therefore, we can simplify
the square root of 4 with 2,
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and we remain with
the square root of 2.
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OK, therefore, what
we can conclude--
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that the two points of
intersections between the two
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graphs are which ones?
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The first is the point 3 plus
the square root of 2 and f of 3
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plus square root of
2, which is equal to g
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of 3 plus the square root of 2.
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They are equal exactly by the
computations that we have done.
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And the point 3 minus
square root of 2, f over 3
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minus square root of
2, which is equal to g
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of 3 minus square root of 2.
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OK, let us conclude
computing these terms.
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We get 3 plus the
square root of 2 here.
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And now the image of 3 plus
the square root of 2 through f
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is 6 times 3 plus
square root of 2
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plus 2, which is 18
plus 2, which is 20,
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plus 6 times square root of 2.
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And here we have 3 minus square
root of 2 and the image of 3
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minus square root
of 2 through f is
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6 times 3 minus square
root of 2 plus 2,
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which is again equal to 20
minus 6 times square root of 2.
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Good.
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These two are the points
in common between the two
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graphs of our functions.
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Ciao.
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