If $x=\frac{(\sqrt{3}+1)}{2}$, then the value of $4x^{3}+2x^{2}-8x+7$ is:
$( A).\ 8$
$( B).\ 10$
$( C).\ 15$
$( D).\ 14$

Given: $x=\frac{(\sqrt{3}+1)}{2}$.

To do: To find the value $4x^{3}+2x^{2}-8x+7$.

Solution:
 

Given $x=\frac{( \sqrt{3}+1)}{2}$

$4x^{3}+2x^{2}-8x+7$

$=4( \frac{( \sqrt{3}+1)}{2})^3+2(\frac{( \sqrt{3}+1)}{2})^2-8( \frac{( \sqrt{3}+1)}{2})+7$

$=4\frac{( \sqrt{3}+1)^3}{8}+2\frac{( \sqrt{3}+1)^2}{4}-8\frac{( \sqrt{3}+1)}{2}+7$

$= \frac{( \sqrt{3}+1)^3}{2}+\frac{( \sqrt{3}+1)^2}{2}-4( \sqrt{3}+1)+7$

As known, $( a+b)^3=1a^3+3a^2b+3ab^2+1b^3$ and $( a+b)^2=1a^2+2ab+1b^2$

$=\frac{( 3\sqrt{3}+9+3\sqrt{3}+1)}{2}+\frac{( 3+2\sqrt{3}+1)}{2}-4( \sqrt{3}+1)+7$

$=\frac{( 6\sqrt{3}+10)}{2}+\frac{( 2\sqrt{3}+4)}{2}-4(\sqrt{3}+1)+7$

$=3\sqrt{3}+5+\sqrt{3}+2-4\sqrt{3}-4+7=10$

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Updated on: 10-Oct-2022

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