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

Given:

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

To do: 

We have to find the value of \( 4 x^{3}+2 x^{2}-8 x+7 \).

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

Therefore,

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

$\Rightarrow 2 x=\sqrt{3}+1$

$\Rightarrow 2 x-1=\sqrt{3}$

Squaring both sides, we get,

$(2 x-1)^{2}=(\sqrt{3})^{2}$

$4 x^{2}-4 x+1=3$

$4 x^{2}-4 x+1-3=0$

$4 x^{2}-4 x-2=0$

$2(2 x^{2}-2 x-1)=0$

$2 x^{2}-2 x-1=0$..........(i)

Therefore,

$4 x^{3}+2 x^{2}-8 x+7=4 x^{3}+6 x^{2}-4x^{2}-6x-2x+10-3$

$=(4 x^{3}-4 x^{2}-2 x)+(6 x^{2}-6 x-3)+10$

$=2x(2 x^{2}-2 x-1)+3(2 x^{2}-2 x-1)+10$

$=2 x \times 0+3 \times 0+10$                    [From (i)]

$=0+0+10$

$=10$

The value of \( 4 x^{3}+2 x^{2}-8 x+7 \) is $10$.

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Updated on: 10-Oct-2022

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