Cubic Roots

Algebra Level 4

Let x = a x=a , y = b y=b and z = c z=c be the solutions of the simultaneous equations x + y + z = − 71 , 4 x + 2 y + z = − 32 , 9 x + 3 y + z = − 27. \begin{aligned} x+y+z &= -71, \\ 4x+2y+z &= -32, \\ 9x+3y+z &= -27. \end{aligned} Then the three roots of the cubic equation t 3 + a t 2 + b t + c = 0 t^3+at^2+bt+c=0 are α \alpha , β \beta and γ \gamma , where α < β < γ \alpha< \beta< \gamma . What is the value of 100 α + 10 β + γ 100\alpha+10\beta+\gamma ?

368 368 257 257 146 146 479 479

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Mahindra Jain by Mahindra Jain

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{ x + y + z = − 71 → ( 1 ) 4 x + 2 y + z = − 32 → ( 2 ) 9 x + 3 y + z = − 27 → ( 3 ) \begin{cases}x+y+z=-71\rightarrow (1)\\4x+2y+z=-32\rightarrow (2)\\9x+3y+z=-27\rightarrow (3)\end{cases} ( 2 ) − ( 1 ) ( 3 ) − ( 1 ) ( 4 x + 2 y + z ) − ( x + y + z ) = − 32 − ( − 71 ) ( 9 x + 3 y + z ) − ( x + y + z ) = − 27 − ( − 71 ) 3 x + y = 39 → ( 4 ) 8 x + 2 y = 44 → 4 x + y = 22 → ( 5 ) \begin{array}{r|r} (2)-(1) & (3)-(1) \\ \hline (4x+2y+z)-(x+y+z)=-32-(-71) & (9x+3y+z)-(x+y+z)=-27-(-71) \\ 3x+y=39\rightarrow (4) & 8x+2y=44 \rightarrow 4x+y=22\rightarrow (5) \end{array} ( 5 ) − ( 4 ) ⟹ ( 4 x + y ) − ( 3 x + y ) = 22 − 39 x = − 17 (5)-(4)\implies (4x+y)-(3x+y)=22-39\\ x=-17 x = − 17 ⟹ 4 x + y = 22 → 4 ( − 17 ) + y = 22 y = 22 + 68 = 90 → y = 90 x=-17\implies 4x+y=22\rightarrow 4(-17)+y=22\\ y=22+68=90\rightarrow y=90 x + y + z = − 71 − 17 + 90 + z = − 71 z = − 71 − 73 = − 144 x+y+z=-71\\ -17+90+z=-71\\ z=-71-73=-144 Hence ( a , b , c ) = ( − 17 , 90 , − 144 ) (a,b,c)=(-17,90,-144) .Hence the cubic equation is: t 3 − 17 t 2 + 90 t − 144 = 0 t^3-17t^2+90t-144=0 By the Rational Root Test, t = 3 t=3 is a solution.Dividing y t − 3 t-3 gives t 2 − 14 t + 48 t^2-14t+48 ,which nicely factors into ( t − 8 ) ( t − 6 ) (t-8)(t-6) .Hence t 3 − 17 t 2 + 90 t − 144 = ( t − 3 ) ( t − 6 ) ( t − 8 ) t^3-17t^2+90t-144=(t-3)(t-6)(t-8) .Since α < β < γ \alpha<\beta<\gamma , α = 3 , β = 6 , γ = 8 \alpha=3,\beta=6,\gamma=8 .Hence, 100 α + 10 β + γ = 100 ( 3 ) + 10 ( 6 ) + 8 = 368 100\alpha+10\beta+\gamma=100(3)+10(6)+8=\boxed{368}

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