Let $\alpha, \beta$ be the roots of the equation $x^2 - \sqrt{2}x + 2 = 0$. Then $\alpha^{14} + \beta^{14}$ is equal to Options: (A) $-64$ (B) $-64\sqrt{2}$ (C) $-128$ (D) $-128\sqrt{2}$

### Analyzing the Quadratic Equation We begin with the quadratic equation: $$x^2 - \sqrt{2}x + 2 = 0$$ Applying the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we substitute the coefficients $a=1$, $b=-\sqrt{2}$, and $c=2$: $$x = \frac{\sqrt{2} \pm \sqrt{(\sqrt{2})^2 - 4(1)(2)}}{2(1)}$$ Simplifying the discriminant, we obtain: $$x = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm i\sqrt{6}}{2}$$ ### The Geometry of the Roots To visualize these roots, we factor out $\sqrt{2}$: $$x = \sqrt{2} \left( \frac{1 \pm i\sqrt{3}}{2} \right)$$ Recognizing the trigonometric form, we identify the roots as: $$x = \sqrt{2} \left( \cos\left(\frac{\pi}{3}\right) \pm i\sin\left(\frac{\pi}{3}\right) \right)$$ Thus, our roots $\alpha$ and $\beta$ are complex numbers with magnitude $\sqrt{2}$ and arguments $\pm \frac{\pi}{3}$. ### The Power of Euler We express the roots using Euler's formula, $\alpha = \sqrt{2} e^{i\frac{\pi}{3}}$ and $\beta = \sqrt{2} e^{-i\frac{\pi}{3}}$. We now calculate $\alpha^{14} + \beta^{14}$: $$\alpha^{14} + \beta^{14} = (\sqrt{2})^{14} \left( e^{i\frac{14\pi}{3}} + e^{-i\frac{14\pi}{3}} \right)$$ Since $(\sqrt{2})^{14} = 2^7 = 128$, the expression simplifies to: $$128 \left( e^{i\frac{14\pi}{3}} + e^{-i\frac{14\pi}{3}} \right)$$ ### The Final Symmetry Using the identity $e^{i\theta} + e^{-i\theta} = 2\cos(\theta)$, we transform the expression: $$128 \times 2 \cos\left(\frac{14\pi}{3}\right) = 256 \cos\left(\frac{14\pi}{3}\right)$$ We simplify the angle $\frac{14\pi}{3}$ by removing full rotations ($4\pi = \frac{12\pi}{3}$): $$\cos\left(\frac{14\pi}{3}\right) = \cos\left(4\pi + \frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$$ Given that $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$, the final result is: $$256 \times \left( -\frac{1}{2} \right) = -128$$ The final answer is **-128**.

JEE Main 2023 (13 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let $α, β$ be the roots of the equation $x^{2} - 2 x + 2 = 0$ . Then $α^{14} + β^{14}$ is equal to

Q: Why can't I just use Newton's Sums for this? It seems faster.

Newton's Sums are indeed powerful! You could define $S_n = \alpha^n + \beta^n$ and use the recurrence $S_n - \sqrt{2}S_{n-1} + 2S_{n-2} = 0$. However, for a power as high as 14, the iterative calculation can be prone to arithmetic errors. The polar form approach provides a more elegant, direct path that avoids the 'brute force' of recursion.

Select Answer:

Visualized Solution

Analyze the Equation $x^{2} - 2 x + 2 = 0$

Given equation: $x^{2} - 2 x + 2 = 0$
Identify coefficients: $a = 1$ , $b = - 2$ , $c = 2$

Apply the Quadratic Formula

Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
Substitution: $x = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( 1 ) ( 2 )}{2 ( 1 )}$

Simplify the Discriminant

$x = \frac{2 \pm 2 - 8}{2}$
$x = \frac{2 \pm - 6}{2}$
Using $i = - 1$ : $x = \frac{2 \pm i 6}{2}$

Factorize the Roots

$x = \frac{2 \pm i 2 3}{2}$
$x = \frac{2 ( 1 \pm i 3 )}{2}$
Rewrite as: $x = 2 (\frac{1 \pm i 3}{2})$

Convert to Euler's Form $r e^{i θ}$

Recall Euler's Form: $e^{i θ} = cos θ + i sin θ$
For $θ = \frac{π}{3}$ : $cos \frac{π}{3} + i sin \frac{π}{3} = \frac{1}{2} + i \frac{3}{2}$
Roots: $α = 2 e^{i \frac{π}{3}}$ and $β = 2 e^{- i \frac{π}{3}}$

Visualize on the Complex Plane

$α$ has magnitude $2$ and argument $\frac{π}{3}$
$β$ has magnitude $2$ and argument $- \frac{π}{3}$

Calculate $α^{14} + β^{14}$

Expression: $α^{14} + β^{14}$
Substitute: $(2 e^{i \frac{π}{3}})^{14} + (2 e^{- i \frac{π}{3}})^{14}$
Simplify magnitudes: $(2)^{14} = (2^{\frac{1}{2}})^{14} = 2^{7} = 128$

Apply Euler's Identity

Factor out $2^{7}$ : $2^{7} (e^{i \frac{14 π}{3}} + e^{- i \frac{14 π}{3}})$
Use identity: $e^{i θ} + e^{- i θ} = 2 cos θ$
Result: $2^{7} \cdot 2 cos (\frac{14 π}{3}) = 2^{8} cos (\frac{14 π}{3})$

Simplify the Angle $\frac{14 π}{3}$

Angle reduction: $\frac{14 π}{3} = \frac{12 π + 2 π}{3} = 4 π + \frac{2 π}{3}$
Periodicity: $cos (4 π + θ) = cos θ$
Simplified term: $2^{8} cos (\frac{2 π}{3})$

Final Computation

Value of $cos \frac{2 π}{3} = - \frac{1}{2}$
Calculation: $256 \cdot (- \frac{1}{2})$
Final Result: $- 128$

00:00 / 00:00

The Sigma Insight: Euler's Form and De Moivre's Theorem

Solution Diagram

Analyzing the Quadratic Equation

We begin with the quadratic equation:

x^{2} - 2 x + 2 = 0

Applying the quadratic formula

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

, we substitute the coefficients

a = 1

b = - 2

, and

c = 2

x = \frac{2 \pm ( 2 ) ^{2} - 4 ( 1 ) ( 2 )}{2 ( 1 )}

Simplifying the discriminant, we obtain:

x = \frac{2 \pm 2 - 8}{2} = \frac{2 \pm i 6}{2}

The Geometry of the Roots

To visualize these roots, we factor out

2

x = 2 (\frac{1 \pm i 3}{2})

Recognizing the trigonometric form, we identify the roots as:

x = 2 (cos (\frac{π}{3}) \pm i sin (\frac{π}{3}))

Thus, our roots

α

and

β

are complex numbers with magnitude

2

and arguments

\pm \frac{π}{3}

The Power of Euler

We express the roots using Euler's formula,

α = 2 e^{i \frac{π}{3}}

and

β = 2 e^{- i \frac{π}{3}}

. We now calculate

α^{14} + β^{14}

α^{14} + β^{14} = (2)^{14} (e^{i \frac{14 π}{3}} + e^{- i \frac{14 π}{3}})

Since

(2)^{14} = 2^{7} = 128

, the expression simplifies to:

128 (e^{i \frac{14 π}{3}} + e^{- i \frac{14 π}{3}})

The Final Symmetry

Using the identity

e^{i θ} + e^{- i θ} = 2 cos (θ)

, we transform the expression:

128 \times 2 cos (\frac{14 π}{3}) = 256 cos (\frac{14 π}{3})

We simplify the angle

\frac{14 π}{3}

by removing full rotations (

4 π = \frac{12 π}{3}

cos (\frac{14 π}{3}) = cos (4 π + \frac{2 π}{3}) = cos (\frac{2 π}{3})

Given that

cos (\frac{2 π}{3}) = - \frac{1}{2}

, the final result is:

256 \times (- \frac{1}{2}) = - 128

The final answer is -128.

Common Traps & FAQs

Why can't I just use Newton's Sums for this? It seems faster.

Newton's Sums are indeed powerful! You could define

S_{n} = α^{n} + β^{n}

and use the recurrence

S_{n} - 2 S_{n - 1} + 2 S_{n - 2} = 0

. However, for a power as high as 14, the iterative calculation can be prone to arithmetic errors. The polar form approach provides a more elegant, direct path that avoids the 'brute force' of recursion.

How do I know if I should use the quadratic formula or just complete the square?

Both are valid! Completing the square is often faster if the coefficients are 'friendly'. Here,

x^{2} - 2 x + 2 = 0

can be written as

(x - \frac{2}{2})^{2} = - 2 + \frac{1}{2} = - \frac{3}{2}

. Both methods lead you to the same complex roots. Choose the one that feels most intuitive to your brain today.

Wait, why does the angle

\frac{14 π}{3}

simplify to

\frac{2 π}{3}

Great catch! Remember that the cosine function is periodic with a period of

2 π

. Since

\frac{14 π}{3} = 4 π + \frac{2 π}{3}

, we are essentially completing two full rotations around the unit circle and landing back at the same position as

\frac{2 π}{3}

. Always look for these 'full circles' to simplify your life!

Is it possible to get a positive answer here?

In this specific problem, the cosine of the resulting angle

cos (\frac{2 π}{3})

is negative, which forces the final result to be negative. If you find yourself getting a positive value, double-check your quadrant logic for the angle

\frac{2 π}{3}

—it lies in the second quadrant where cosine is strictly negative.

Do I really need Euler's form, or can I just expand the binomial?

Expanding a binomial to the 14th power is a recipe for disaster! Euler's form is the 'shortcut' that turns exponentiation into simple multiplication of the argument. It is the difference between doing 14 lines of algebra and 2 lines of elegant trigonometry.

Animated Solution for Mathematics - Complex Numbers: Let $α, β$ be the roots of the equation $x^{2} - 2 x + 2 = 0$ . Then $α^{14} + β^{14}$ is equal to

Select Answer:

Visualized Solution

Analyze the Equation $x^{2} - 2 x + 2 = 0$

Apply the Quadratic Formula

Simplify the Discriminant

Factorize the Roots

Convert to Euler's Form $r e^{i θ}$

Visualize on the Complex Plane

Calculate $α^{14} + β^{14}$

Apply Euler's Identity

Simplify the Angle $\frac{14 π}{3}$

Final Computation

The Sigma Insight: Euler's Form and De Moivre's Theorem

Analyzing the Quadratic Equation

The Geometry of the Roots

The Power of Euler

The Final Symmetry

Similar Questions

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α,β be the roots of the equation x2−2​x+2=0. Then α14+β14 is equal to

Select Answer:

Visualized Solution

Analyze the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−2​x+2=0

Apply the Quadratic Formula

Simplify the Discriminant

Factorize the Roots

Convert to Euler's Form .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } reiθ

Visualize on the Complex Plane

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Apply Euler's Identity

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Final Computation

The Sigma Insight: Euler's Form and De Moivre's Theorem

Analyzing the Quadratic Equation

The Geometry of the Roots

The Power of Euler

The Final Symmetry

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Animated Solution for Mathematics - Complex Numbers: Let $α, β$ be the roots of the equation $x^{2} - 2 x + 2 = 0$ . Then $α^{14} + β^{14}$ is equal to

Analyze the Equation $x^{2} - 2 x + 2 = 0$

Convert to Euler's Form $r e^{i θ}$

Calculate $α^{14} + β^{14}$

Simplify the Angle $\frac{14 π}{3}$

Let $α, β$ be the roots of the quadratic equation $x^{2} + 6 x + 3 = 0$ . Then $\frac{α ^{23} + β ^{23} + α ^{14} + β ^{14}}{α ^{15} + β ^{15} + α ^{10} + β ^{10}}$ is equal to

If $(3 + i)^{100} = 2^{99} (p + i q)$ , then $p$ and $q$ are roots of the equation :

Find all non-zero complex numbers $Z$ satisfying $\overset{ˉ}{Z} = i Z^{2}$ .

Let $S = {a + b 2 : a, b \in Z}$ , $T_{1} = {(- 1 + 2)^{n} : n \in Z}$ , and $T_{2} = {(1 + 2)^{n} : n \in N}$ . Then which of the following statements is(are) TRUE ?

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If $z = \frac{3}{2} + \frac{i}{2}, i = - 1$ , then $(z^{201} - i)^{8}$ is equal to

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The value of $(\frac{1 + s i n \frac{2 π}{9} + i c o s \frac{2 π}{9}}{1 + s i n \frac{2 π}{9} - i c o s \frac{2 π}{9}})^{3}$ is