Roots From Other Dimension

Algebra Level 4

Let x 1 x_1 , x 2 x_{2} and x 3 x_{3} be the roots of x 3 + a x 2 + a x + 1 = 0 x^{3}+ax^{2}+ax+1=0 and y 1 y_1 , y 2 y_{2} and y 3 y_{3} be the roots of y 3 + 2 a y 2 + 2 a y + 1 = 0 y^{3}+2ay^{2}+2ay+1=0 , where a > 0 a>0 . If y 1 2 + y 2 2 + y 3 2 = x 1 2 + x 2 2 + x 3 2 + 133 y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+133 , find a a .


The answer is 7.

Watsapol Kerdlarppol by Watsapol Kerdlarppol , 22, Thailand

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1 solution

By Vieta's formula , we have:

{ x 1 + x 2 + x 3 = − a x 1 x 2 + x 2 x 3 + x 3 x 1 = a \begin{cases} x_1+x_2+x_3 = - a \\ x_1x_2+x_2x_3+x_3x_1 = a \end{cases}

⟹ x 1 2 + x 2 2 + x 3 2 = ( x 1 + x 2 + x 3 ) 2 − 2 ( x 1 x 2 + x 2 x 3 + x 3 x 1 ) = ( − a ) 2 − 2 a = a 2 − 2 a \begin{aligned} \implies x_1^2+x_2^2+x_3^2 & = (x_1+x_2+x_3)^2 - 2(x_1x_2+x_2x_3+x_3x_1) \\ & = (-a)^2 - 2a \\ & = a^2 - 2a \end{aligned}

And

{ y 1 + y 2 + y 3 = − 2 a y 1 y 2 + y 2 y 3 + y 3 y 1 = 2 a \begin{cases} y_1+y_2+y_3 = - 2a \\ y_1y_2+y_2y_3+y_3y_1 = 2a \end{cases}

⟹ y 1 2 + y 2 2 + y 3 2 = ( y 1 + y 2 + y 3 ) 2 − 2 ( y 1 y 2 + y 2 y 3 + y 3 y 1 ) = ( − 2 a ) 2 − 2 ( 2 a ) = 4 a 2 − 4 a \begin{aligned} \implies y_1^2+y_2^2+y_3^2 & = (y_1+y_2+y_3)^2 - 2(y_1y_2+y_2y_3+y_3y_1) \\ & = (-2a)^2 - 2(2a) \\ & = 4a^2 - 4a \end{aligned}

Therefore,

y 1 2 + y 2 2 + y 3 2 = x 1 2 + x 2 2 + x 3 2 + 133 ⟹ 4 a 2 − 4 a = a 2 − 2 a + 133 3 a 2 − 2 a − 133 = 0 ( a − 7 ) ( 3 a + 19 ) = 0 ⟹ a = 7 for a > 0 \begin{aligned} y_1^2+y_2^2+y_3^2 & = x_1^2+x_2^2+x_3^2 + 133 \\ \implies 4a^2-4a & = a^2-2a+133 \\ 3a^2 - 2a - 133 & = 0 \\ (a-7)(3a+19) & = 0 \\ \implies a & = \boxed{7} & \text{for }a > 0 \end{aligned}

4 Helpful 0 Interesting 1 Brilliant 0 Confused

in the last line it should be ( a − 7 ) ( 3 a + 19 ) = 0 (a-7)(3a+19)=0 :)

Anirudh Sreekumar - 4 years, 2 months ago

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Thanks, I have changed it.

Chew-Seong Cheong - 4 years, 2 months ago

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