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Question

Mathematics Question on Complex Numbers and Quadratic Equations

If 2α=1i32 \alpha =-1-i \sqrt{3} and 2β=1+i32 \beta=-1+i \sqrt {3}, then 5α4+5β4+7α1β15 \alpha^4+5 \beta^4 +7 \alpha ^{-1} \beta ^{-1} is equal to

A

1- 1

B

2-2

C

22

D

11

Answer

22

Explanation

Solution

Given that, 2α=1i32\alpha=-1-i \sqrt{3} and 2β=1+i32\beta=-1+i\sqrt{3} α+β=1\therefore \alpha+\beta=-1 and αβ=1\alpha\beta=1 Now, 5α4+5β4+7αβ5\alpha^{4}+5\beta^{4}+\frac{7}{\alpha\beta} =5\left[\left\\{\left(\alpha+\beta\right)^{2}-2\alpha\beta^{3}-2\left(\alpha\beta\right)\right\\}^{2}\right]+\frac{7}{\alpha\beta} = 5\left[\left\\{\left(-1\right)^{2}-2\times1\right\\}^{2}-2\left(1\right)^{2}\right]+\frac{7}{1} =5(12)+7=2=5\left(1-2\right)+7=2.