We're asked to find a cubic equation
with roots 1/2, negative 1/3, and two
using Vieta's formula.
Vieta's formula relates the roots of a polynomial
to its coefficients.
For a cubic equation, if we let alpha, beta, and gamma
be the roots of this cubic equation,
then the sum and product of the roots
are given by these three equations
where sum one equals S sub one,
which equals alpha plus beta plus gamma,
which equals negative B divided by A,
and sum two equals S sub two,
which equals alpha times beta
plus beta times gamma, plus gamma times alpha,
which equals C divided by A,
and finally the product P equals alpha times beta
times gamma, which equals negative D divided by A.
Once we find these two sums and the product,
we can use S sub one, S sub two, and P
in this equation here to determine a cubic equation
with roots alpha, beta, and gamma,
or we can go back to these equations
and determine A, B, C, and D
and use this form of a cubic equation.
In this video we will show both methods.
To keep things organized let's set alpha equal 1/2,
beta equal negative 1/3, and gamma equal two.
Let's begin by determining S sub one.
S sub one is equal to,
again, alpha plus beta plus gamma,
which is 1/2 plus negative 1/3
plus two.
Let's write plus two as plus two over a one,
and now we obtain a common denominator
to determine the sum.
The least common denominator is six.
We write each fraction as an equivalent fraction
with the denominator of six.
Notice now we have a common nominator of six,
and the numerator is three plus negative two
plus 12, which gives us 13/6.
Now we find S sub two or sum two,
which is equal to alpha times beta,
which is 1/2 times negative 1/3,
plus beta times gamma, which is plus negative 1/3
times two or times two over one,
and then plus gamma times alpha,
which is two times 1/2.
Now let's determine the products.
Here we have negative 1/6
plus negative 2/3 or minus 2/3.
Then here we have plus two times 1/2,
which gives us plus one.
Let's write that as plus one over one,
and now we must obtain a common nominator.
The least common nominator is six.
So now we write each fraction as an equivalent fraction
with the denominator of six.
Now we have a common nominator of six,
and the numerator is negative one minus four
plus six, which gives us positive 1/6,
and now we need to find the product, P,
which is equal to alpha times beta times gamma,
which is 1/2 times negative 1/3 times two
or times two over one.
Simplifying before multiplying,
there's one two and two here and here,
and now multiplying we have negative 1/3.
So now that we have S sub one, S sub two, and P,
we can use this formula here
to determine a possible cubic equation
with the given roots.
So we have X cubed
minus S sub one X squared.
S sub one is 13/6, which gives us minus 13/6 X squared.
Then we have plus S sub two X.
S sub two is 1/6 giving us plus 1/6 X,
and then we have minus P.
Because P is negative 1/3,
we have minus negative 1/3,
which is equivalent to plus 1/3.
And this is equal to zero.
So this is a possible cubic equation
that has the given roots,
but let's also find an equivalent cubic equation
that doesn't contain fractions.
Notice the least common nominator is six,
and therefore we can determine an equivalent equation
without fractions by multiplying both sides by six.
Let's do this on the next slide.
So we multiply the left side by six
and the right side by six.
Six times X cubed is six X cubed
minus six times 13/6 is 13
giving us minus 13 X squared,
six times 1/6 is one giving us plus one X,
or just plus X, and six times 1/3 is two
giving us plus two.
On the right side we still have zero.
So it's not wrong to use the equation with fractions,
but this equation might be more common,
which brings up another important point,
there are an infinite number of cubic equations
that have these three roots.
We can determine as many equivalent equations
as we wish by multiplying both sides
of the equation by any constant.
Now let's go back and determine the equation
of a cubic by determining the values of A, B, C, and D
from the two sums and the product.
So again we already know that S sub one
is equal to 13/6,
but this must also equal negative B divided by A,
and notice how I put the negative in the numerator,
which will be helpful.
And S sub two is equal to 1/6,
which must also equal C over A or C divided by A,
and finally the product P
is equal to negative 1/3 or negative one over three,
which is equal to negative D divided by A.
Notice how for these three fractions
the denominators are all A,
which means all three of these fractions
must have the same denominator,
and these two fractions already have denominator of six,
so let's multiply the numerator and denominator
of negative 1/3 by two so it also has a denominator of six.
Now we can say the product P
is equal to negative two divided by six,
which again is equal to negative D divided by A.
So now we can disregard this equation here,
and because all the denominators must be A,
we can see that A must be equal to six.
Looking at this equation we know A is six,
which means negative B must be 13,
and therefore B is equal to negative 13.
Looking at this equation, again, because we know
A is six, C is equal to one,
and finally looking at this equation,
because A is equal to six, negative D equals negative two,
and therefore D equals positive two.
So now we can use the values of A, B, C, and D
and substitute into this cubic equation
to determine the cubic equation
with the given roots.
Performing this substitution
we have six X cubed,
and then plus B X squared
would be plus negative 13 X squared,
or just minus 13 X squared,
and then plus C X is plus one X or just plus X,
and then plus D is just plus two,
and this is equal to zero.
So notice how this is the same equation we found earlier
when we used this formula with S sub one, S sub two, and P.
Again, this equation is the same equation
we found earlier, which is here.
I hope you found this helpful.