Question

Let x=√(4-√7) and y = √(4+√7)
then
((x-y)/√3)^2+(x^2y+xy^2)^2 +(x+y)^2
(A) 328
(B) O
(C)956/9
(D)433/9

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Answer 1

Answer:

I know the answer , this question is of Allen , the final answer is D) 422/3 with no doubt

Answer 2

D) 422/3 is the answer.

Step-by-step explanation:

Given: $x = \sqrt{4 - \sqrt{7}}$ and $y = \sqrt{4 + \sqrt{7}}$

To Find: $\Big(\dfrac{x-y}{\sqrt{3}} \Big)^{2} + (x^2y+xy^2)^2 + (x+y)^2$

Solution:

• To evaluate $\Big(\dfrac{x-y}{\sqrt{3}} \Big)^{2} + (x^2y+xy^2)^2 + (x+y)^2$

Simplifying the expression $\Big(\dfrac{x-y}{\sqrt{3}} \Big)^{2} + (x^2y+xy^2)^2 + (x+y)^2$ such that we have,

$\Rightarrow \Big(\dfrac{x-y}{\sqrt{3}} \Big)^{2} + (x^2y+xy^2)^2 + (x+y)^2$

$\Rightarrow \Big(\dfrac{x-y}{\sqrt{3}} \Big)^{2} + (xy)^{2} (x+y)^2 + (x+y)^2$

$\Rightarrow \Big(\dfrac{x-y}{\sqrt{3}} \Big)^{2} + ((xy)^{2} + 1 ) (x+y)^2$

$\Rightarrow \Big(\dfrac{x^{2} + y^{2} - 2xy}{3} \Big) + (x^{2}y^{2} + 1 ) (x^{2}+y^{2} + 2xy)$

Now, substituting $x = \sqrt{4 - \sqrt{7}}$ and $y = \sqrt{4 + \sqrt{7}}$ in the above expression, we get,

$\Rightarrow \Big(\dfrac{4 - \sqrt{7} + 4 + \sqrt{7} - 2(\sqrt{4 - \sqrt{7}})(\sqrt{4 + \sqrt{7}})}{3} \Big) + (({4 - \sqrt{7}})({4 - \sqrt{7}}) + 1 ) (({4 - \sqrt{7}})+({4 + \sqrt{7}}) + 2(\sqrt{4 - \sqrt{7}})(\sqrt{4 + \sqrt{7}}))$

$\Rightarrow \Big(\dfrac{8 - 2(4^{2} - (\sqrt{7})^{2})^{1/2}}{3} \Big) + ((4^{2} - (\sqrt{7})^{2})^{1/2} + 1 ) (8 + 2(4^{2} - (\sqrt{7})^{2})^{1/2})$

$\Rightarrow \Big(\dfrac{8 - 2(16 - 7)^{1/2}}{3} \Big) + (16 - 7 + 1 ) (8 + 2(16 - 7)^{1/2})$

$\Rightarrow \Big(\dfrac{8 - 2(9)^{1/2}}{3} \Big) + (10) (8 + 2(9)^{1/2})$

$\Rightarrow \Big(\dfrac{8 - 2(3)}{3} \Big) + (10) (8 + 2(3))$

$\Rightarrow \Big(\dfrac{2}{3} \Big) + (10) (14) = \dfrac{422}{3}$

Hence, 422/3 is the answer.