Roots above the ground?

Algebra Level 3

α , β , γ , δ \alpha, \beta, \gamma, \delta are the roots of the equation x 4 a x 3 + a x 2 + b x + c = 0 \large\ { x }^{ 4 } - a{ x }^{ 3 } + a{ x }^{ 2 } + bx + c = 0 where a , b , c a, b, c are real numbers. What is the smallest possible value of α 2 + β 2 + γ 2 + δ 2 \large\ { \alpha }^{ 2 } + { \beta }^{ 2 } + { \gamma }^{ 2 } + { \delta }^{ 2 } ?


The answer is -1.

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1 solution

Applying Vieta-Cardano formulas: α 2 + β 2 + γ 2 + δ 2 = ( α + β + γ + δ ) 2 2 ( α β + α γ + α δ + β γ + β δ + γ δ ) = a 2 2 a \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta) = a^2 - 2a \Rightarrow The smallest value of α 2 + β 2 + γ 2 + δ 2 \alpha^2 + \beta^2 + \gamma^2 + \delta^2 is attained at the vertex v v of the parabola a 2 2 a = a ( a 2 ) a^2 - 2a = a(a - 2) which vertes is v = 1 v = 1 ,i.e,

the smallest value of α 2 + β 2 + γ 2 + δ 2 \alpha^2 + \beta^2 + \gamma^2 + \delta^2 is the smallest value of a 2 2 a a^2 - 2a which is 1 \boxed{-1} attained at the vertex v = 1 v = 1 of this parabola

How can sum of 4 squares give -1? The appropriate answer should be 0.

Yash Gadhia - 4 years, 9 months ago

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because there may be complex roots....

Guillermo Templado - 4 years, 9 months ago

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