It is actually very simple. To solve an equation of the form:
x^3 + a x^2 + b x + c = 0
you follow the following steps.
First, get rid of the quadratic term
a x^2 using the substitution:
x = y - a/3
This step is analogous to how you solve the quadratic equation when you write it as a perfect square.
In this case we aren't finshed yet as the equation now is of the form:
y^3 + p y + q = 0
How do we solve this equation?
Consider the identity:
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
We can rewrite the right hand side as follows:
a^3 + 3a^2b + 3ab^2 + b^3 =
a^3 + b^3 + 3ab(a+b)
So, we have:
(a+b)^3 = 3ab(a+b)+ a^3 + b^3 ---->
(a+b)^3 - 3ab(a+b) - [a^3+b^3] = 0
Of course, this equation is always satisfied whatever values for a and b you substitute in it. Now, if we choose a and b such that:
3ab = -p (1)
a^3 + b^3 = -q (2)
Then:
(a+b)^3 +p(a+b) + q = 0
so y = a + b is then the solution we are looking for.
So, it all boils down to solving equations (1) and (2) for a and b.
If you thake the third power of (1):
27 a^3b^3 = -p^3
So, if we put A = a^3 and B = B^3, we have:
A*B = -p^3/27
A + B = -q
If you eliminate, say, B , you get a quadratic equation for A which you know how to solve. You then only need to pay attention when you extract the cube roots to find a and b. Equation (1) has to be satisfied, which means that if you choose one of the three possible complex cube roots for A, the root for B is fixed.
How do I algebraically find the x-intercepts for this equation y = x^3 - 3x^2 + 3.
I know I need to plug in y = 0 to solve for x.
x^3 - 3x^2 + 3 = 0
But where do I go from there? I don't think I can factor, and I don't think I can use the quadratic formula.
5 answers
So, let's see how this works out for in your case:
x^3 - 3x^2 + 3 = 0
We put x = y + 1, which gives after some simple algebra:
y^3 - 3y + 1 = 0
So, p = -3 and q = 1
We need to solve:
A*B = -p^3/27 = 1
A + B = -q = -1
The second equation says that
B = -(1+A)
Substituting in the first gives:
-A^2 - A = 1 --->
A^2 + A + 1 = 0 ----->
A = -1/2 ± i sqrt[3]/2
We have two solutions because interchanging A and B will give you another solution. But if you do that the other solution for A will be the first solution for B. So we can take:
A = -1/2 + i sqrt[3]/2
B = -1/2 - i sqrt[3]/2
We can rewrite this as:
A = cos(4 pi/3) - i sin(4 pi/3)
B = cos(4 pi/3) + i sin(4 pi/3)
Which we can also write as:
A = exp(-4 pi i/3)
B = exp(4 pi i/3)
Which means that:
a = exp(-4 pi i/9 + 2 pi i n/3)
b = exp(4 pi i/9 - 2 pi i n/3)
Note that the product of a and b must be 1, the integer n which selects which of the three complex cube roots we choose for A fixes the cube root for B.
The solution is thus:
y = a + b =
exp(-4 pi i/9 + 2 pi i n/3)
+ exp(4 pi i/9 - 2 pi i n/3) =
2 cos(4 pi/9 + 2 pi n/3)
And
x = y + 1 = 2 cos(4 pi/9 + 2 pi n/3)
The three solutions are thus:
x = 2 cos(4 pi/9) + 1
x = 2 cos(10 pi/9) + 1
x = 2 cos(2 pi/9) + 1
x^3 - 3x^2 + 3 = 0
We put x = y + 1, which gives after some simple algebra:
y^3 - 3y + 1 = 0
So, p = -3 and q = 1
We need to solve:
A*B = -p^3/27 = 1
A + B = -q = -1
The second equation says that
B = -(1+A)
Substituting in the first gives:
-A^2 - A = 1 --->
A^2 + A + 1 = 0 ----->
A = -1/2 ± i sqrt[3]/2
We have two solutions because interchanging A and B will give you another solution. But if you do that the other solution for A will be the first solution for B. So we can take:
A = -1/2 + i sqrt[3]/2
B = -1/2 - i sqrt[3]/2
We can rewrite this as:
A = cos(4 pi/3) - i sin(4 pi/3)
B = cos(4 pi/3) + i sin(4 pi/3)
Which we can also write as:
A = exp(-4 pi i/3)
B = exp(4 pi i/3)
Which means that:
a = exp(-4 pi i/9 + 2 pi i n/3)
b = exp(4 pi i/9 - 2 pi i n/3)
Note that the product of a and b must be 1, the integer n which selects which of the three complex cube roots we choose for A fixes the cube root for B.
The solution is thus:
y = a + b =
exp(-4 pi i/9 + 2 pi i n/3)
+ exp(4 pi i/9 - 2 pi i n/3) =
2 cos(4 pi/9 + 2 pi n/3)
And
x = y + 1 = 2 cos(4 pi/9 + 2 pi n/3)
The three solutions are thus:
x = 2 cos(4 pi/9) + 1
x = 2 cos(10 pi/9) + 1
x = 2 cos(2 pi/9) + 1
I guess we will have to find one of the roots by brute force.
We are looking for zeros of f(x) for some x
Make a table to try to find the zeros
for example
x f(x)
0 +3
1 +1
2 -1
3 +3
That is interesting, f(x) goes through zero at least once between x = +1 and x +2. Try x = 1.3
1.3 +1.27
1.4 -.136
Oh, so between 1.3 and 1.4
Try 1.33
1.33 +.045937
1.34 +.019304
1.35 -.007125
Humm, I am not going do go any further. That is close enough for now. The function crosses the x axis at x is about 1.35
Now what. I need to factor that root out so I am left with a quadratic or else continue graphing and calculating to find the other two roots.
To factor it out, note that (x-1.35) is a factor (to our level of approximation) so
(x-1.35)(a x^2 + b x +c) = x^3-3x^2+3
so we need to divide the right by (x-1.35)
I get
x^2 - 1.65 x - 2.2275
solve that quadratic to find the other two roots.
However I bet the problem is a typo unless you have covered this stuff in class.
We are looking for zeros of f(x) for some x
Make a table to try to find the zeros
for example
x f(x)
0 +3
1 +1
2 -1
3 +3
That is interesting, f(x) goes through zero at least once between x = +1 and x +2. Try x = 1.3
1.3 +1.27
1.4 -.136
Oh, so between 1.3 and 1.4
Try 1.33
1.33 +.045937
1.34 +.019304
1.35 -.007125
Humm, I am not going do go any further. That is close enough for now. The function crosses the x axis at x is about 1.35
Now what. I need to factor that root out so I am left with a quadratic or else continue graphing and calculating to find the other two roots.
To factor it out, note that (x-1.35) is a factor (to our level of approximation) so
(x-1.35)(a x^2 + b x +c) = x^3-3x^2+3
so we need to divide the right by (x-1.35)
I get
x^2 - 1.65 x - 2.2275
solve that quadratic to find the other two roots.
However I bet the problem is a typo unless you have covered this stuff in class.
Wow... that's a long process. Thanks for the help!
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1 answer