Let α, β be the roots of the equation x² - ax - b = 0 with Im(α)

Q: Why can't we just solve for $\alpha$ and $\beta$ using the quadratic formula?

That is a great instinct, but look at the coefficients $a$ and $b$. They are unknowns! If you try to use the quadratic formula, you will be dealing with square roots of expressions involving $a$ and $b$, which makes raising them to the 6th power a nightmare. The recurrence relation method is the 'elegant' path—it bypasses the need to know the roots individually and works directly with the power sums.

Q: Wait, why does $P_{n} = aP_{n-1} + bP_{n-2}$ work?

Think of it this way: since $\alpha$ is a root, $\alpha^{2} = a\alpha + b$. Multiply both sides by $\alpha^{n-2}$, and you get $\alpha^{n} = a\alpha^{n-1} + b\alpha^{n-2}$. The same holds for $\beta$. When you subtract the equation for $\beta$ from the equation for $\alpha$, the terms group perfectly into $P_{n}$, $P_{n-1}$, and $P_{n-2}$. It is a direct consequence of the definition of the roots.

Q: How do I know which identity to use for $|\alpha^{4} + \beta^{4}|$?

Whenever you have the difference of two quantities ($P_{4} = \alpha^{4} - \beta^{4}$) and you need the sum ($\alpha^{4} + \beta^{4}$), always look for the bridge between them. The identity $(X+Y)^{2} = (X-Y)^{2} + 4XY$ is that bridge. By setting $X = \alpha^{4}$ and $Y = \beta^{4}$, you transform a problem about sums into a problem about differences and products, which we have already solved!

Question

Let $\alpha, \beta$ be the roots of the equation $x^{2}-ax-b=0$ with $Im(\alpha)<Im(\beta)$. Let $P_{n}=\alpha^{n}-\beta^{n}$. If $P_{3}=-5\sqrt{7}i, P_{4}=-3\sqrt{7}i, P_{5}=11\sqrt{7}i$ and $P_{6}=45\sqrt{7}i,$ then $|\alpha^{4}+\beta^{4}|$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the DNA of the Quadratic

Every quadratic equation carries the 'DNA' of its roots. Since $\alpha$ and $\beta$ are roots of $x^{2}-ax-b=0$, they must satisfy the equation. This means:

$$\alpha^{2} - a\alpha - b = 0$$
$$\beta^{2} - a\beta - b = 0$$

If we multiply the first by $\alpha^{n-2}$ and the second by $\beta^{n-2}$, we obtain:

$$\alpha^{n} - a\alpha^{n-1} - b\alpha^{n-2} = 0$$
$$\beta^{n} - a\beta^{n-1} - b\beta^{n-2} = 0$$

Subtracting these two equations is where the magic happens. We arrive at the recurrence relation:

$$P_{n} = aP_{n-1} + bP_{n-2}$$

This relation is our master key. It allows us to relate the higher powers of the roots to the lower ones without ever needing to calculate the roots themselves.

### The System of Equations

Now, we put our master key to work using the given values for $P_{6}, P_{5}, P_{4},$ and $P_{3}$. For $n=6$, we have $P_{6} = aP_{5} + bP_{4}$. Substituting the given values:

$$45\sqrt{7}i = a(11\sqrt{7}i) + b(-3\sqrt{7}i)$$

Notice how the $\sqrt{7}i$ term appears in every single term. We can divide it out completely, leaving us with the clean linear equation:

$$45 = 11a - 3b$$

We repeat this for $n=5$, using $P_{5} = aP_{4} + bP_{3}$, which gives:

$$11\sqrt{7}i = a(-3\sqrt{7}i) + b(-5\sqrt{7}i)$$

Again, the $\sqrt{7}i$ vanishes, leaving $11 = -3a - 5b$. We now have a simple system of two linear equations:

$$11a - 3b = 45$$
$$3a + 5b = -11$$

Solving this system is straightforward. Multiplying the first by $5$ and the second by $3$, we get:

$$55a - 15b = 225$$
$$9a + 15b = -33$$

Adding them yields $64a = 192$, so $a = 3$. Substituting $a=3$ back into the second equation, we find $3(3) + 5b = -11$, which simplifies to $5b = -20$, so $b = -4$. The quadratic equation is revealed:

$$x^{2} - 3x - 4 = 0$$

### The Algebraic Bridge

We have successfully identified the quadratic equation. Now, we need to find $|\alpha^{4} + \beta^{4}|$. We know the difference $P_{4} = \alpha^{4} - \beta^{4} = -3\sqrt{7}i$.

We use the classic algebraic identity $(X+Y)^{2} = (X-Y)^{2} + 4XY$. By setting $X = \alpha^{4}$ and $Y = \beta^{4}$, we get:

$$(\alpha^{4} + \beta^{4})^{2} = (P_{4})^{2} + 4(\alpha\beta)^{4}$$

From Vieta's formulas, the product of the roots $\alpha\beta$ is the constant term of $x^{2} - 3x - 4 = 0$, which is $-4$. Now, we calculate the components:

First, $(P_{4})^{2} = (-3\sqrt{7}i)^{2} = 9 \cdot 7 \cdot i^{2} = 63 \cdot (-1) = -63$.

Second, $4(\alpha\beta)^{4} = 4 \cdot (-4)^{4} = 4 \cdot 256 = 1024$.

Adding these together:

$$(\alpha^{4} + \beta^{4})^{2} = -63 + 1024 = 961$$

Finally, taking the square root, we find:

$$|\alpha^{4} + \beta^{4}| = \sqrt{961} = \mathbf{31}$$

We have traversed the path from a complex recurrence relation to a simple, elegant integer. The beauty of mathematics lies in this very ability to simplify the complex.

Animated Solution for Mathematics - Quadratic Equations: Let $α, β$ be the roots of the equation $x^{2} - a x - b = 0$ with $I m (α) < I m (β)$ . Let $P_{n} = α^{n} - β^{n}$ . If $P_{3} = - 57 i, P_{4} = - 37 i, P_{5} = 117 i$ and $P_{6} = 457 i,$ then $∣ α^{4} + β^{4} ∣$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Equation $x^{2} - a x - b = 0$

Deriving the Recurrence Relation $P_{n}$

Applying Recurrence for $n = 6$

Applying Recurrence for $n = 5$

Solving for $a$ and $b$

Finding the Product of Roots $α β$

The Algebraic Identity for $∣ α^{4} + β^{4} ∣$

Calculating the Square of $P_{4}$

Calculating the Product Term $4 (α β)^{4}$

Final Summation and Result

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the DNA of the Quadratic

The System of Equations

The Algebraic Bridge

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Comprehension Passage

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Enter Numerical Value:

Visualized Solution

Understanding the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−ax−b=0

Deriving the Recurrence Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pn​

Applying Recurrence for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=6

Applying Recurrence for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=5

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Finding the Product of Roots .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } αβ

The Algebraic Identity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣α4+β4∣

Calculating the Square of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P4​

Calculating the Product Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4(αβ)4