Question:
If $x=\frac{\sqrt{3}+1}{2}$, find the value of $4 x^{3}+2 x^{2}-8 x+7$
Solution:
Given, $x=\frac{\sqrt{3}+1}{2}$, and given to find the value of $4 x^{3}+2 x^{2}-8 x+7$
$2 x=\sqrt{3}+1$
$2 x-1=\sqrt{3}$
Now, squaring on both the sides, we get, $(2 x-1)^{2}=3$
$4 x^{2}-4 x+1=3$
$4 x^{2}-4 x+1-3=0$
$4 x^{2}-4 x-2=0$
$2 x^{2}-2 x-1=0$
Now taking $4 x^{3}+2 x^{2}-8 x+7$
$2 x\left(2 x^{2}-2 x-1\right)+4 x^{2}+2 x+2 x^{2}-8 x+7$
$2 x\left(2 x^{2}-2 x-1\right)+6 x^{2}-6 x+7$
As, $2 x^{2}-2 x-1=0$
$\left.2 x(0)+3\left(2 x^{2}-2 x-1\right)\right)+7+3$
0 + 3(0) + 10
10
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