Let $\alpha, \beta$ be the roots of the equation $x^2 + 2\sqrt{2}x - 1 = 0$. The quadratic equation, whose roots are $\alpha^4 + \beta^4$ and $\frac{1}{10}(\alpha^6 + \beta^6)$, is : Options: (A) $x^2 - 190x + 9466 = 0$ (B) $x^2 - 180x + 9506 = 0$ (C) $x^2 - 195x + 9506 = 0$ (D) $x^2 - 195x + 9466 = 0$

### Analyzing the Setup We are given the quadratic equation $x^2 + 2\sqrt{2}x - 1 = 0$ with roots $\alpha$ and $\beta$. Our objective is to construct a new quadratic equation with roots $R_1 = \alpha^4 + \beta^4$ and $R_2 = \frac{1}{10}(\alpha^6 + \beta^6)$. Instead of using the quadratic formula, which introduces cumbersome radicals, we utilize **Vieta's formulas**. For the given equation, the sum and product of the roots are: $$\alpha + \beta = -2\sqrt{2}$$ $$\alpha\beta = -1$$ ### The Ladder of Powers We calculate the powers of the roots systematically. First, we find $\alpha^2 + \beta^2$ using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$: $$\alpha^2 + \beta^2 = (-2\sqrt{2})^2 - 2(-1) = 8 + 2 = 10$$ Next, we determine $R_1 = \alpha^4 + \beta^4$ by applying the identity $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2$: $$R_1 = (10)^2 - 2(-1)^2 = 100 - 2 = 98$$ ### The Leap to the Sixth Power To find $\alpha^6 + \beta^6$, we treat it as a sum of cubes: $(\alpha^2)^3 + (\beta^2)^3$. Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ where $a = \alpha^2$ and $b = \beta^2$, we have: $$\alpha^6 + \beta^6 = (\alpha^2 + \beta^2)(\alpha^4 - \alpha^2\beta^2 + \beta^4)$$ Substituting our known values $\alpha^2 + \beta^2 = 10$, $\alpha^4 + \beta^4 = 98$, and $\alpha\beta = -1$: $$\alpha^6 + \beta^6 = 10 \times (98 - (-1)^2) = 10 \times 97 = 970$$ Given $R_2 = \frac{1}{10}(\alpha^6 + \beta^6)$, we find: $$R_2 = \frac{970}{10} = 97$$ ### The Final Construction We now have the roots of our new quadratic equation: $R_1 = 98$ and $R_2 = 97$. The equation is given by $x^2 - (\text{Sum})x + (\text{Product}) = 0$. The sum of the roots is: $$R_1 + R_2 = 98 + 97 = 195$$ The product of the roots is: $$R_1 \times R_2 = 98 \times 97 = 9506$$ Thus, the resulting quadratic equation is: **$x^2 - 195x + 9506 = 0$**

Why can't we just find $\alpha$ and $\beta$ using the quadratic formula? It seems more direct.

While tempting, solving for $\alpha$ and $\beta$ introduces messy radicals that make calculating $\alpha^6$ and $\beta^6$ a nightmare. Symmetric polynomials allow us to work with integers, keeping the path clean and error-free.

I got stuck on $\alpha^6 + \beta^6$. Is there a specific formula?

Think of $\alpha^6 + \beta^6$ as $(\alpha^2)^3 + (\beta^2)^3$. Use the sum of cubes identity: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, where $a = \alpha^2$ and $b = \beta^2$. It turns a complex power into a simple multiplication problem.

How do I know if I've made a calculation error?

Always check the symmetry. If your intermediate steps for $\alpha^n + \beta^n$ don't result in simple integers, re-check your Vieta's coefficients. In JEE problems, these values are usually designed to be clean.

JEE Main 2024 (09 Apr Shift 1)

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Animated Solution for Mathematics - Quadratic Equations: Let $α, β$ be the roots of the equation $x^{2} + 22 x - 1 = 0$ . The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6})$ , is :

Select Answer:

Visualized Solution

The Original Equation

Given quadratic equation: $x^{2} + 22 x - 1 = 0$
Let the roots be $α$ and $β$ .
Our goal is to find a new equation with roots $R_{1} = α^{4} + β^{4}$ and $R_{2} = \frac{1}{10} (α^{6} + β^{6})$ .

Vieta's Formulas

Sum of roots: $α + β = - \frac{b}{a}$
Product of roots: $α β = \frac{c}{a}$

Extracting Sum and Product

$α + β = - \frac{2 2}{1} = - 22$
$α β = - \frac{1}{1} = - 1$

Stepping Stone: $α^{2} + β^{2}$

We need higher powers, so we first find $α^{2} + β^{2}$ .
Algebraic identity: $α^{2} + β^{2} = (α + β)^{2} - 2 α β$

Calculating $α^{2} + β^{2}$

Substitute the values: $α^{2} + β^{2} = (- 22)^{2} - 2 (- 1)$
$(- 22)^{2} = 8$
$- 2 (- 1) = + 2$
$α^{2} + β^{2} = 8 + 2 = 10$

First New Root: $R_{1}$

First new root: $R_{1} = α^{4} + β^{4}$
Identity: $α^{4} + β^{4} = (α^{2} + β^{2})^{2} - 2 (α β)^{2}$

Calculating $R_{1}$

Substitute $α^{2} + β^{2} = 10$ and $α β = - 1$
$R_{1} = (10)^{2} - 2 (- 1)^{2}$
$R_{1} = 100 - 2 (1)$
$R_{1} = 98$

Path to the Second Root

We need $α^{6} + β^{6}$ for the second root.
Sum of cubes identity: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$
Let $a = α^{2}$ and $b = β^{2}$
$α^{6} + β^{6} = (α^{2} + β^{2}) (α^{4} - α^{2} β^{2} + β^{4})$

Calculating $α^{6} + β^{6}$

Rearrange: $α^{6} + β^{6} = (α^{2} + β^{2}) [(α^{4} + β^{4}) - (α β)^{2}]$
Substitute values: $10 \times [98 - (- 1)^{2}]$
$10 \times [98 - 1] = 10 \times 97$
$α^{6} + β^{6} = 970$

Second New Root: $R_{2}$

Second new root: $R_{2} = \frac{1}{10} (α^{6} + β^{6})$
Substitute the calculated value: $R_{2} = \frac{1}{10} (970)$
$R_{2} = 97$

Forming the New Equation

A quadratic equation with roots $R_{1}$ and $R_{2}$ is given by:
$x^{2} - (Sum of roots) x + (Product of roots) = 0$
$x^{2} - (R_{1} + R_{2}) x + (R_{1} \cdot R_{2}) = 0$

Sum and Product of New Roots

Sum of new roots: $S = R_{1} + R_{2} = 98 + 97 = 195$
Product of new roots: $P = R_{1} \times R_{2} = 98 \times 97$
$P = (100 - 2) (100 - 3) = 10000 - 300 - 200 + 6 = 9506$

The Final Quadratic Equation

Substitute $S = 195$ and $P = 9506$ into the template.
Final Equation: $x^{2} - 195 x + 9506 = 0$
This matches Option 2.

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The Sigma Insight: Relation Between Roots and Coefficients

Solution Diagram

Analyzing the Setup

We are given the quadratic equation

x^{2} + 22 x - 1 = 0

with roots

α

and

β

. Our objective is to construct a new quadratic equation with roots

R_{1} = α^{4} + β^{4}

and

R_{2} = \frac{1}{10} (α^{6} + β^{6})

Instead of using the quadratic formula, which introduces cumbersome radicals, we utilize Vieta's formulas. For the given equation, the sum and product of the roots are:

α + β = - 22

α β = - 1

The Ladder of Powers

We calculate the powers of the roots systematically. First, we find

α^{2} + β^{2}

using the identity

α^{2} + β^{2} = (α + β)^{2} - 2 α β

α^{2} + β^{2} = (- 22)^{2} - 2 (- 1) = 8 + 2 = 10

Next, we determine

R_{1} = α^{4} + β^{4}

by applying the identity

α^{4} + β^{4} = (α^{2} + β^{2})^{2} - 2 (α β)^{2}

R_{1} = (10)^{2} - 2 (- 1)^{2} = 100 - 2 = 98

The Leap to the Sixth Power

To find

α^{6} + β^{6}

, we treat it as a sum of cubes:

(α^{2})^{3} + (β^{2})^{3}

. Using the identity

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

where

a = α^{2}

and

b = β^{2}

, we have:

α^{6} + β^{6} = (α^{2} + β^{2}) (α^{4} - α^{2} β^{2} + β^{4})

Substituting our known values

α^{2} + β^{2} = 10

α^{4} + β^{4} = 98

, and

α β = - 1

α^{6} + β^{6} = 10 \times (98 - (- 1)^{2}) = 10 \times 97 = 970

Given

R_{2} = \frac{1}{10} (α^{6} + β^{6})

, we find:

R_{2} = \frac{970}{10} = 97

The Final Construction

We now have the roots of our new quadratic equation:

R_{1} = 98

and

R_{2} = 97

. The equation is given by

x^{2} - (Sum) x + (Product) = 0

The sum of the roots is:

R_{1} + R_{2} = 98 + 97 = 195

The product of the roots is:

R_{1} \times R_{2} = 98 \times 97 = 9506

Thus, the resulting quadratic equation is:

$x^{2} - 195 x + 9506 = 0$

Animated Solution for Mathematics - Quadratic Equations: Let $α, β$ be the roots of the equation $x^{2} + 22 x - 1 = 0$ . The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6})$ , is :

Select Answer:

Visualized Solution

The Original Equation

Vieta's Formulas

Extracting Sum and Product

Stepping Stone: $α^{2} + β^{2}$

Calculating $α^{2} + β^{2}$

First New Root: $R_{1}$

Calculating $R_{1}$

Path to the Second Root

Calculating $α^{6} + β^{6}$

Second New Root: $R_{2}$

Forming the New Equation

Sum and Product of New Roots

The Final Quadratic Equation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Ladder of Powers

The Leap to the Sixth Power

The Final Construction

Similar Questions

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Select Answer:

Visualized Solution

The Original Equation

Vieta's Formulas

Extracting Sum and Product

Stepping Stone: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2

First New Root: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Path to the Second Root

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α6+β6

Second New Root: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Forming the New Equation

Sum and Product of New Roots

The Final Quadratic Equation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Ladder of Powers

The Leap to the Sixth Power

The Final Construction

Similar Questions

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Stepping Stone: $α^{2} + β^{2}$

Calculating $α^{2} + β^{2}$

First New Root: $R_{1}$

Calculating $R_{1}$

Calculating $α^{6} + β^{6}$

Second New Root: $R_{2}$

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