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Question : 94 of 160
Marks:
+1,
-0
Solution:
Given quadratic equation is
x2−2x+4=0
whose roots are α and β.
∴ α+β=2 and αβ=4 ....…(i)
Now,
α9+β9=(α3)3+(β3)3=(α3+β3) (α6+β6−α3β3)
=(α+β)(α2−αβ+β2)[(α2)3+(β2)3−α3β3]
=(α+β)[(α+β)2−3αβ] [(α2+β2)(α4+β4−α2β2)−α3β3]
=(α+β)[(α+β)2−3αβ][{(α+β)2−2αβ} {(α2+β2)2−3α2β2}−α3β3]
=(α+β)[(α+β)2−3αβ][{(α+β)2−2αβ} [{(α+β)2−(2αβ)]2−3α2β2}−α3β3]
=2[4−12][{4−8}{(4−8)2−48}−64] [from Eq. (i)]
=2(−8){(−4)(−32)(−64)}
=2(−8)(128−64)
=2(−8)(64)=−210
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