x^( 2) − 4x −1=0 α , β are roots α^( 3) + 17β +5 =? −−−solution−−− α is root ⇒ α^( 2) −4α −1=0 ⇒ α^( 2) = 4α +1 ✓ α^( 3) + 17β +5 = α . α^( 2) +17β +5 = α ( 4α +1 )+ 17β +5 = 4α^( 2) + α + 17β +5 = 4 (4α +1 )+α +17β +5=17(α+β)+9 = 17S +9= 17 (4 )+9=77


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Question Number 162174 by mnjuly1970 last updated on 27/Dec/21

$x^{2} - 4 x - 1 = 0$ $α, β a r e r o o t s$ $α^{3} + 17 β + 5 = ?$ $- - - s o l u t i o n - - -$ $α i s r o o t \Rightarrow α^{2} - 4 α - 1 = 0$ $\Rightarrow α^{2} = 4 α + 1 ✓$ $α^{3} + 17 β + 5 = α . α^{2} + 17 β + 5$ $= α (4 α + 1) + 17 β + 5$ $= 4 α^{2} + α + 17 β + 5$ $= 4 (4 α + 1) + α + 17 β + 5 = 17 (α + β) + 9$ $= 17 S + 9 = 17 (4) + 9 = 77$
Commented by cortano last updated on 27/Dec/21
$x^2 −4x−1=0 { (α),(β) :}⇒α+β=4 ⇒α^2 = 4α+1 ⇒ α^3 = 4α^2 +α ⇒α^3 +17β+5 = 4α^2 +α+17β+5 = 4(4α+1)+16β+9 = 16(α+β)+13 = 64+13 = 77$
$x^{2} - 4 x - 1 = 0 {\begin{cases} α \\ β \end{cases} \Rightarrow α + β = 4$ $\Rightarrow α^{2} = 4 α + 1$ $\Rightarrow α^{3} = 4 α^{2} + α$ $\Rightarrow α^{3} + 17 β + 5 = 4 α^{2} + α + 17 β + 5$ $= 4 (4 α + 1) + 16 β + 9$ $= 16 (α + β) + 13$ $= 64 + 13 = 77$
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