Question

If x=5 - root 3 upon 5 + root 3 and x= 5 + root 3 upon 5 - root 3, show that x square - y square = 10 root 3 upon 11

Answer 1

Hey friend, Harish here.

Here is your answer:

Given that,

1) $x = \frac{5- \sqrt{3} }{5+ \sqrt{3} }$

2) $y = \frac{5+ \sqrt{3} }{5- \sqrt{3} }$

To prove:

$x^{2} -y^{2} = \frac{10 \sqrt{3} }{11}$

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,

$x = \frac{5- \sqrt{3} }{5+ \sqrt{3} }\times \frac{5- \sqrt{3} }{5- \sqrt{3} } = \frac{(5- \sqrt{3} )^{2}}{(5+ \sqrt{3})(5- \sqrt{3}) }$

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

⇒ $x = \frac{(5- \sqrt{3})^{2} }{22} = \frac{(25+3-10 \sqrt{3}) }{22} = \frac{28-10 \sqrt{3} }{22} = \frac{2(14-5 \sqrt{3}) }{2\times 11} = \frac{14-5 \sqrt{3} }{11}$

Now rationalize y using the same method.

⇒ $y = \frac{(5+ \sqrt{3})^{2} }{22} = \frac{(25+3+10 \sqrt{3}) }{22} = \frac{28+10 \sqrt{3} }{22} = \frac{2(14+5 \sqrt{3}) }{2\times 11} = \frac{14+5 \sqrt{3} }{11}$

Now we know that,

x² - y² = (x + y) (x - y)

⇒ $(\frac{(14+5 \sqrt{3})+(14-5 \sqrt{3}) }{11})( \frac{(14+5 \sqrt{3})-(14-5 \sqrt{3})}{11})$

⇒  $( \frac{28}{11})( \frac{10 \sqrt{3}}{11}) = \frac{280 \sqrt{3} }{121}$

$\bold{Therefore \ the\ answer\ is\ \ \ \boxed{ \frac{280 \sqrt{3} }{121} } }$
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Hope my answer is helpful to you.

Answer 2

Answer:

Hence, the value of:

$x^2-y^2=\dfrac{-280\sqrt{3}}{121}$

Step-by-step explanation:

If:

$x=\dfrac{5-\sqrt{3}}{5+\sqrt{3}}$

and

$y=\dfrac{5+\sqrt{3}}{5-\sqrt{3}}$

Now we are asked to find the value of:

$x^2-y^2$

We know that:

$x^2-y^2=(x-y)(x+y)$

so,

$x-y=\dfrac{5-\sqrt{3}}{5+\sqrt{3}}-\dfrac{5+\sqrt{3}}{5-\sqrt{3}}\\\\\\x-y=\dfrac{(5-\sqrt{3})(5-\sqrt{3})-(5+\sqrt{3})(5+\sqrt{3})}{(5-\sqrt{3})(5+\sqrt{3})}\\\\x-y=\dfrac{(5-\sqrt{3})^2-(5+\sqrt{3})^2}{25-3}\\\\x-y=\dfrac{25+3-10\sqrt{3}-25-3-10\sqrt{3}}{22}\\\\x-y=\dfrac{-20\sqrt{3}}{22}\\\\x-y=\dfrac{-10\sqrt{3}}{11}$

Similarly,

$x+y=\dfrac{5-\sqrt{3}}{5+\sqrt{3}}+\dfrac{5+\sqrt{3}}{5-\sqrt{3}}\\\\\\x+y=\dfrac{(5-\sqrt{3})(5-\sqrt{3})+(5+\sqrt{3})(5+\sqrt{3})}{(5-\sqrt{3})(5+\sqrt{3})}\\\\\\x+y=\dfrac{(5-\sqrt{3})^2+(5+\sqrt{3})^2}{25-3}\\\\x+y=\dfrac{25+3-10\sqrt{3}+25+3+10\sqrt{3}}{22}\\\\x+y=\dfrac{56}{22}\\\\x+y=\dfrac{28}{11}$

Hence,

$x^2-y^2=\dfrac{28}{11}\times \dfrac{-10\sqrt{3}}{11}\\\\x^2-y^2=\dfrac{-280\sqrt{3}}{121}$

Hence, the value of:

$x^2-y^2=\dfrac{-280\sqrt{3}}{121}$