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NARRATOR: Exercise five.

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For what values of k
does the equation

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(4k + 1) times x squared minus
2(k + 1) times x + (1 – 2k) = 0

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have one real solution?

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For what values of k, if any,
is the quadratic negative definite?

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When considering the number of
solutions of a quadratic equation

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of the form ax squared + bx + c = 0,

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we must consider the discriminant,
b squared – 4ac.

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In the case of this quadratic,
a is equal to (4k + 1),

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b is equal to –2(k + 1)

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and c is equal to (1 – 2k).

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So the discriminant
can be calculated as

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(-2(k + 1)) all squared,
minus 4(4k + 1) times (1 – 2k).

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And when we expand and simplify,

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we find that the discriminant

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is equal to 36k squared.

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A quadratic has only one solution
when its discriminant is 0.

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In this case that is
when 36k squared is equal to 0,

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and this will happen when k = 0.

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So therefore, this quadratic equation
has one real solution

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when k is equal to 0.

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A quadratic is negative definite

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if the graph of the quadratic
sits entirely below the x-axis.

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So this means that
the y-coordinate of the vertex

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would need to be negative

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and the parabola
would need to be inverted

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so that the coefficient
of x squared is negative.

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And also we would note that the

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quadratic would have no x-intercepts.

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In the case of this quadratic,
we know that the discriminant -

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which tells us about the number
of solutions of the equation

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and hence the number of
x-intercepts of the graph -

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this discriminant is 36k squared,

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which is always greater than
or equal to 0,

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no matter the value of k.

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So if the discriminant
cannot be negative,

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this quadratic will always
have x-intercepts

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and hence it cannot
be negative definite.