Question : If $y=1+\sqrt{3}+\sqrt{4}$, then the value of $2 y^4-8 y^3-6 y^2+28 y-84$ is:
Option 1: $40 \sqrt{3}$
Option 2: $80 \sqrt{3}$
Option 3: $20 \sqrt{3}$
Option 4: $60 \sqrt{3}$
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Correct Answer: $40 \sqrt{3}$
Solution :
$y=1+\sqrt{3}+\sqrt{4}=1+\sqrt{3}+2= 3+\sqrt{3}$ -----------(i)
Now, $2 y^4-8 y^3-6 y^2+28 y-84$
= $2y^2(y^2-4y-3)+28y-84$
= $2y^2(y^2-4y+4-4-3)+28y-84$
= $2y^2((y-2)^2-7)+28y-84$
Substituting the value of $y$ from (i),
$2y^2((y-2)^2-7)+28y-84$
= $2(3+\sqrt{3})^2[(3+\sqrt{3}-2)^2-7]+28(3+\sqrt{3})-84$
= $2(9+6\sqrt{3}+3)[(1+2\sqrt{3}+3)-7]+84+28\sqrt{3}-84$
= $(24+12\sqrt{3})(2\sqrt{3}-3)+28\sqrt{3}$
= $48\sqrt{3}+72-72-36\sqrt{3}+28\sqrt{3}$
= $40\sqrt{3}$
Hence, the correct answer is $40\sqrt{3}$.
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