Question

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

• A. $0$
• B. $1$
• C. $10$
• D. 7

Answer 1

The correct option is C $10$

Given that,

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

$⟹2x=\sqrt{3}+1$

$⟹2x-1=\sqrt{3}$

$⟹(2x-1{)}^2=3$

$⟹4x^2-4x+1=3$...........(i)

$⟹4x^2=4x+3-1$

$⟹4x^2=4x+2⟹2x^2=2x+1$.......(ii)

Multiplying with $x$ on both the sides, we get

$⟹4x^3=4x^2+2x$..............(iii)

$\therefore 4x^3+2x^2-8x+7$

$=(4x^2+2x)+(2x+1)-8x+7$

$=4x^2+4x-8x+1+7$

$=4x^2-4x+1+7$

$=3+7$

$=10$(Ans)

Option C is correct.