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Answer:
Hey friend, Harish here.
Here is your answer:
Given that,
1) x = \frac{5- \sqrt{3} }{5+ \sqrt{3} }x=
5+
3
5−
3
2) y = \frac{5+ \sqrt{3} }{5- \sqrt{3} }y=
5−
3
5+
3
To prove:
x^{2} -y^{2} = \frac{10 \sqrt{3} }{11}x
2
−y
2
=
11
10
3
Solution:
First we must rationalize the denominators of x & y.
To rationalize the denominators we must multiply and divide the number by it's conjugate.
Conjugate of x is 5 - √3.
Conjugate of y is 5 +√3.
Then,
x = \frac{5- \sqrt{3} }{5+ \sqrt{3} }\times \frac{5- \sqrt{3} }{5- \sqrt{3} } = \frac{(5- \sqrt{3} )^{2}}{(5+ \sqrt{3})(5- \sqrt{3}) }x=
5+
3
5−
3
×
5−
3
5−
3
=
(5+
3
)(5−
3
)
(5−
3
)
2
Now for multiplying the denominators use the identity (a+b)(a-b) = a² - b².
Then, (5 + √3) (5 - √3) = 5² - (√3)² = 25 - 3 = 22
⇒ x = \frac{(5- \sqrt{3})^{2} }{22} = \frac{(25+3-10 \sqrt{3}) }{22} = \frac{28-10 \sqrt{3} }{22} = \frac{2(14-5 \sqrt{3}) }{2\times 11} = \frac{14-5 \sqrt{3} }{11}x=
22
(5−
3
)
2
=
22
(25+3−10
3
)
=
22
28−10
3
=
2×11
2(14−5
3
)
=
11
14−5
3
Now rationalize y using the same method.
⇒ y = \frac{(5+ \sqrt{3})^{2} }{22} = \frac{(25+3+10 \sqrt{3}) }{22} = \frac{28+10 \sqrt{3} }{22} = \frac{2(14+5 \sqrt{3}) }{2\times 11} = \frac{14+5 \sqrt{3} }{11}y=
22
(5+
3
)
2
=
22
(25+3+10
3
)
=
22
28+10
3
=
2×11
2(14+5
3
)
=
11
14+5
3
Now we know that,
x² - y² = (x + y) (x - y)
⇒ (\frac{(14+5 \sqrt{3})+(14-5 \sqrt{3}) }{11})( \frac{(14+5 \sqrt{3})-(14-5 \sqrt{3})}{11})(
11
(14+5
3
)+(14−5
3
)
)(
11
(14+5
3
)−(14−5
3
)
)
⇒ ( \frac{28}{11})( \frac{10 \sqrt{3}}{11}) = \frac{280 \sqrt{3} }{121}(
11
28
)(
11
10
3
)=
121
280
3
\bold{Therefore \ the\ answer\ is\ \ \ \boxed{ \frac{280 \sqrt{3} }{121} } }Therefore the answer is
121
280
3
_______________________________________________
Hope my answer is helpful to you.
Step-by-step explanation:
