Dont mess with k.. ok!!

Algebra Level 2

If k = 3 + 1 2 \frac{\sqrt{3}+1}{2} Then find the value of 4 k 3 k^{3} + 2 k 2 k^{2} - 8k + 7


The answer is 10.

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2 solutions

Uahbid Dey
Jun 17, 2014

k=½(√3+1) k² = 1+½√3 => 2k² = 2+√3 k³ = k.k² = ¼(5+3√3) => 4k³ = 5+3√3 =>....... 4k³ = 5+3√3 +2k² = 2+√3 -8k = -4√3 - 4 +7 = +7 ---------------------- .....= 10

forgot to divide by 4 for the 2k^2 just couldn't see where I was going wrong!

Filly Mare - 6 years, 12 months ago
Sanjeet Raria
Oct 27, 2014

We have k = 3 + 1 2 k=\frac{√3+1}{2} 2 k 1 = 3 \Rightarrow 2k-1=√3 Squaring & solving, we get 2 k 2 2 k 1 = 0 \Rightarrow 2k^2-2k-1=0 Now dividing the required expression 4 k 3 + 2 k 2 8 k + 7 4k^3+2k^2-8k+7 by 2 k 2 2 k 1 2k^2-2k-1 we get 4 k 3 + 2 k 2 8 k + 7 = ( 2 k 2 2 k 1 ) ( 2 k + 3 ) + 10 4k^3+2k^2-8k+7=(2k^2-2k-1)(2k+3)+10 4 k 3 + 2 k 2 8 k + 7 = 0 + 10 = 10 \Rightarrow 4k^3+2k^2-8k+7=0+10=\boxed{10}

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