Question

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

• A. $0$
• B. $100$
• C. $11$
• D. $10$

Answer 1

The correct option is D $10$

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

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

On squaring both sides, we get,

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

$\Rightarrow 4x^2=4x+2$

Also, $\Rightarrow 2x^2=2x+1$ -----$\left(i\right)$

On multiplying $\left(i\right)$ by $x$ both sides, we get,

$4x^3=4x^2+2x$

$4x^3=4x+2+2x=6x+2$ (Substituting $\left(i\right)$)

$4x^3=6x+2$ -----$\left(ii\right)$

Then, $4x^3+2x^2-8x+7$

$=(6x+2)+(2x+1)-8x+7$ (Substitute $\left(i\right)$ and $\left(ii\right)$)

$6x+2x-8x+2+1+7=8x-8x+10=10$.

Hence, Option $D$ is correct.