Its all about the roots!

Algebra Level 3

Given that α \alpha and β \beta are the real roots of x 2 − 2 x − 1 = 0 , x^2-2x-1=0, find the value of 5 α 4 + 12 β 3 5\alpha^4+12\beta^3 .


The answer is 169.

Sathvik Acharya by Sathvik Acharya , 17, India

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1 solution

Chew-Seong Cheong
Aug 29, 2017

Relevant wiki: Vieta's Formula - Quadratics - Basic

x 2 − 2 x − 1 = 0 ⟹ x 2 = 2 x + 1 x 3 = 2 x 2 + x = 2 ( 2 x + 1 ) + x ⟹ x 3 = 5 x + 2 . . . ( 1 ) x 4 = 5 x 2 + 2 x = 5 ( 2 x + 1 ) + 2 x ⟹ x 4 = 12 x + 5 . . . ( 2 ) \begin{aligned} x^2 - 2x-1 & = 0 \\ \implies x^2 & = 2x+1 \\ x^3 & = 2{\color{#3D99F6}x^2} + x = 2{\color{#3D99F6}(2x+1)} + x \\ \implies x^3 & = 5x+2 & ... (1) \\ x^4 & = 5{\color{#3D99F6}x^2} + 2x = 5{\color{#3D99F6}(2x+1)} + 2x \\ \implies x^4 & = 12x+5 & ...(2) \end{aligned}

Therefore we have:

5 α 4 + 12 β 3 = 5 ( 12 α + 5 ) ⏟ ( 2 ) + 12 ( 5 β + 2 ) ⏟ ( 1 ) = 60 α + 25 + 60 β + 24 = 60 ( α + β ) + 49 By Vieta’s formula α + β = 2 = 60 ( 2 ) + 49 = 169 \begin{aligned} 5{\color{#3D99F6}\alpha^4} + 12{\color{#D61F06}\beta^3} & = 5\underbrace{\color{#3D99F6}(12\alpha + 5)}_{(2)} + 12\underbrace{\color{#D61F06}(5\beta+2)}_{(1)} \\ & = 60 \alpha + 25 + 60 \beta + 24 \\ & = 60({\color{#3D99F6}\alpha + \beta}) + 49 & \small \color{#3D99F6} \text{By Vieta's formula }\alpha + \beta = 2 \\ & = 60({\color{#3D99F6}2}) + 49 \\ & = \boxed{169} \end{aligned}

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