In this video I go over the second step in deriving the Cubic Formula, and this involves converting our earlier PQ version of the cubic formula into a quadratic formula using Vieta's substitution. In step 1 we converted x into y by the equation x = y + k. In step 2 we convert y into z by the equation y = z + k/z. Then we do the exact same procedure as before to select a k value that cancels out the z1 terms. Then, multiplying by z3 we obtain a quadratic equation, which we can solve using the PQ version of the quadratic formula. In later steps, we will convert z back into y and then finally back into x.
Timestamps
- Step 2: Solve PQ Cubic Equation using Vieta's substitution to obtain quadratic formula: 0:00
- Let y = z + k/z: 0:57
- Select k to cancel out z terms: 3:07
- k = -P/3: 4:13
- Convert resulting equation into a quadratic equation by multiplying by z3: 6:54
- Recall PQ version of the Quadratic Formula: 7:54
- Apply the PQ quadratic formula to our equation: 9:51
- Solution for z3: 10:58
Notes and playlists
- Summary:
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- Notes: https://peakd.com/hive-128780/@mes/dzekfnxh .
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Part 1/5:
Solving Cubic Equations Using V's Substitution
In mathematics, solving cubic equations can often be quite complex. However, applying V's substitution offers a systematic approach to simplify this process. In this article, we will delve into the method of transforming a cubic equation of the form ( y^3 + py + Q = 0 ) into a simpler quadratic equation, enabling us to find the roots of the cubic more easily.
Understanding the PQ Cubic Equation
Part 2/5:
The cubic equation in question can be structured into two components: a linear term ( py ) and a constant ( Q ). The challenge lies in resolving the ( y^3 ) term alongside the other components. To facilitate this, we can utilize V's substitution effectively. By substituting ( y ) with ( z + \frac{K}{z} ), we introduce a new variable that alters the form of the equation.
Applying the Substitution
Let’s define ( y = z + \frac{K}{z} ). Substituting this into the cubic equation results in a complex expression that needs to be expanded. Applying the binomial theorem, particularly using Pascal's triangle, allows us to expand ( (z + \frac{K}{z})^3 ) efficiently. The expansion yields terms that include ( z^3 ), ( z^2 ), and constants involving ( K ).
Part 3/5:
Simplifying the Equation
The goal at this stage is to eliminate the ( pz ) terms. By organizing and combining the resulting coefficients, we aim to cancel out specific terms to streamline the equation. Upon achieving the necessary cancellations, we can determine the value of ( K ) in relation to the coefficient ( p ). It is discovered that:
[
K = -\frac{p}{3}
]
This determination allows further simplification to occur.
Replacing and Rearranging
After finding ( K ), we substitute it back into the transformed cubic expression. The terms reconfigure to eliminate much complexity:
[
-z^3 + \frac{p^3}{27} z^3 + Q = 0
]
Part 4/5:
After resolving and collecting like terms, the equation can be reduced to a standard quadratic form, enabling the application of the quadratic formula.
Deriving the Quadratic Formula
To solve the quadratic equation, we recall the standard form:
[
x^2 + Px + Q = 0
]
Using the quadratic formula yields:
[
x = \frac{-P \pm \sqrt{P^2 - 4Q}}{2}
]
In our case, the variables convert so that ( z^3 = -\frac{p}{2} ), ( P = \frac{p^2}{4} ), and ( Q = \frac{p^3}{27} ). By substituting these values back, we can clearly express ( z^3 ) in terms of ( p ) and ( Q ), leading us to the final roots of the original cubic equation.
Conclusion
Part 5/5:
Through the method of V's substitution, we transform a complicated cubic equation into a manageable quadratic form, thereby revealing solutions with much greater ease than direct calculation methods. This systematic approach not only simplifies the computations but also deeply enhances our understanding of the relationships within polynomial equations. Employing substitution techniques is valuable not only in mathematics but also in various scientific applications requiring precise computational methods.
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