Question
If $\text{x}=\frac{\sqrt3+1}{2},$ find the value of $4\text{x}^3+2\text{x}^2-8\text{x}+7.$
✓
Answer
We have, $\text{x}=\frac{\sqrt3+1}{2}$ It can be simplified as $2\text{x}-1=\sqrt3$ On squaring both sides,
we get $(2\text{x}-1)^2=\big(\sqrt3\big)^2$
$(2\text{x})^2+1-2\times2\text{x}=3$
$4\text{x}^2+1-4\text{x}=3$
$4\text{x}^2-4\text{x}-2=0$
The given equation can be rewritten as. $4\text{x}^3+2\text{x}^2-8\text{x}+7$
$\ =\text{x}(4\text{x}^2-4\text{x}-2)+\frac{6}{4}(4\text{x}^2-4\text{x}-2)+3+7$
Therefore, we have $4\text{x}^3+2\text{x}^2-8\text{x}+7=\text{x}(0)+\frac{6}{4}(0)+3+7$
$=3+7$
$=10$
Hence, the value of given expression is $10.$
we get $(2\text{x}-1)^2=\big(\sqrt3\big)^2$
$(2\text{x})^2+1-2\times2\text{x}=3$
$4\text{x}^2+1-4\text{x}=3$
$4\text{x}^2-4\text{x}-2=0$
The given equation can be rewritten as. $4\text{x}^3+2\text{x}^2-8\text{x}+7$
$\ =\text{x}(4\text{x}^2-4\text{x}-2)+\frac{6}{4}(4\text{x}^2-4\text{x}-2)+3+7$
Therefore, we have $4\text{x}^3+2\text{x}^2-8\text{x}+7=\text{x}(0)+\frac{6}{4}(0)+3+7$
$=3+7$
$=10$
Hence, the value of given expression is $10.$
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