Question

if x=root3-1/root3+1,y=root3+1/root3-1,simplify

Answer 1

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YOUR QUESTION:

if X=√3-1/√3+1  Y=√3+1/√3-1,simplify

ANSWER:


X=(√3+1)2/3-1

=3+1+2√3/2

=4+2√3/2

=2(2+√3)/2

X=2+√3


now,

Y=√3-1/√3+1

=√3-1/√3+1.√3-1/√3-1

=(√3-1)2/3-1

=3+1-2√3/2

=4-2√3/2

=2(2-√3)/2

now,

=X²+Y²+XY

=(2+√3)2+(2-√3)2+(2+√3)(2-√3)

=4+3+√3+4+3-4√3+4-3

=7+7+1

=15

THEREFORE,the answer is 15 




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Answer 2

Answer is 15

Step-by-step explanation:

Given, x = $\frac {\sqrt {3} -1}{\sqrt {3} +1}$ and y = $\frac {\sqrt {3} +1}{\sqrt {3} -1}$

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

On factorizing we get:

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

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

x =$\frac {3^2 + 1^2 -2\sqrt 3}{(\sqrt {3})^2 - (1)^2}$ (using $(a-b)^2 = a^2 - 2ab + b^2$)

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

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

x = $\frac {4-2\sqrt 3}{2}$

x = $2-\sqrt3$

Now for y = $\frac {\sqrt {3} +1}{\sqrt {3} -1}$

On factorizing we get:

y = $\frac {\sqrt {3} +1}{\sqrt {3} -1}$

y = $\frac {(\sqrt {3} +1)^2}{(\sqrt {3})^2 - (1)^2}$

y = $\frac {3^2 + 1^2 +2\sqrt 3}{(\sqrt {3})^2 - (1)^2}$  (using $(a+b)^2 = a^2 + 2ab + b^2$)

y = $\frac {3+1+2\sqrt3}{2}$

y = $\frac {4+2\sqrt 3}{2}$

y = 2+$\sqrt 3$

Now, For $x^2+ y^2 + xy$, we have to put the values of x & y calculated above-

$(2-\sqrt3)^2 + (2+\sqrt3)^2 + (2-\sqrt3)(2+\sqrt3)$

= $4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 + 4 - 3$ using the algebraic identities - (using $(a-b)^2 = a^2 - 2ab + b^2$,  

(using $(a+b)^2 = a^2 + 2ab + b^2$) and $(a+b)(a-b) = a^2 - b^2$

= 14 + 1

= 15 Answer!