Solve x^3 -3x^2 +5x +4 = 0 using Tartaglia's method.
To solve the equation x^3 - 3x^2 + 5x + 4 = 0 using Tartaglia's method, we can follow these steps:
Step 1: Divide the equation by the leading coefficient.
Our equation is already in standard form, so we don't need to perform this step.
Step 2: Make a substitution.
Let's substitute x = y - (b / 3a), where a, b, and c are the coefficients of x^2, x, and the constant term, respectively. In this case, a = 1, b = -3, and c = 5.
Substituting these values:
x = y - (b / 3a)
x = y - (-3 / (3 * 1))
x = y + 1
Now, re-write the equation in terms of y:
(y + 1)^3 - 3(y + 1)^2 + 5(y + 1) + 4 = 0
Step 3: Expand and simplify the equation.
Expanding the equation:
(y + 1)^3 - 3(y + 1)^2 + 5(y + 1) + 4 = 0
(y^3 + 3y^2 + 3y + 1) - 3(y^2 + 2y + 1) + 5y + 5 + 4 = 0
y^3 + 3y^2 + 3y + 1 - 3y^2 - 6y - 3 + 5y + 9 = 0
y^3 - y + 7 = 0
Step 4: Solve the equation y^3 - y + 7 = 0 using Tartaglia's Method.
Tartaglia's method involves observing that y = 1 is one solution of the equation and dividing the equation by (y - 1), which is the corresponding factorization of the polynomial.
Dividing the equation y^3 - y + 7 = 0 by (y - 1):
(y^3 - y + 7) / (y - 1) = 0
We can then use synthetic division to divide the polynomial:
1 | 1 0 -1 7
| 1 1 0
--------------
1 1 0 7
The result of synthetic division is 1y^2 + 1y + 7.
Step 5: Solve the quadratic equation 1y^2 + 1y + 7 = 0.
Since we have a quadratic equation, we can use the quadratic formula to find the solutions. The quadratic formula is given by:
y = [-b ± √(b^2 - 4ac)] / (2a)
In this case, a = 1, b = 1, and c = 7. Substituting these values into the quadratic formula:
y = [-1 ± √(1^2 - 4(1)(7))] / (2 * 1)
y = [-1 ± √(1 - 28)] / 2
y = [-1 ± √(-27)] / 2
Since the discriminant (b^2 - 4ac) is negative, the quadratic equation has complex solutions. Let's simplify the expression further:
y = [-1 ± √(27)i] / 2
So the solutions of the quadratic equation 1y^2 + 1y + 7 = 0 in terms of y are:
y = (-1 + √(27)i) / 2
y = (-1 - √(27)i) / 2
Step 6: Find the solutions for x from the solutions for y.
To find the solutions for x, we use the relation x = y + 1 (from the substitution we made in Step 2).
Substituting y = (-1 + √(27)i) / 2:
x = (-1 + √(27)i) / 2 + 1
x = (-1 + √(27)i) / 2 + 2/2
x = (-1 + 2 + √(27)i) / 2
x = (1 + √(27)i) / 2
Substituting y = (-1 - √(27)i) / 2:
x = (-1 - √(27)i) / 2 + 1
x = (-1 - √(27)i) / 2 + 2/2
x = (-1 - 2 - √(27)i) / 2
x = (-3 - √(27)i) / 2
So the two complex solutions for x are:
x = (1 + √(27)i) / 2
x = (-3 - √(27)i) / 2
Therefore, the equation x^3 - 3x^2 + 5x + 4 = 0 has two complex solutions using Tartaglia's method.