Question

If X equals to 5 minus under root 3 upon 5 + under root 3 and Y = 5 + under root 3 upon 5 minus under root 3 show that AC square minus y squared equals to minus 10 under root 3 upon 11

It is ques 18

Answer 1

See the attachment for the solution.

Answer 2

Answer:

$Now,\\x^{2}-y^{2}\\=\frac{-280\sqrt{3}}{121}$

Step-by-step explanation:

$Given \\x=\frac{5-\sqrt{3}}{5+\sqrt{3}}\\and\: y=\frac{5+\sqrt{3}}{5-\sqrt{3}}$

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

$=\frac{2[5^{2}+\left(\sqrt{3}\right)^{2}]}{(5^{2}-\left(\sqrt{3}\right)^{2}}$

/* By algebraic identities:

i) (a-b)²+(a+b)² = 2(a²+b²)

ii)(a+b)(a-b) = a²-b² */

$=\frac{2(25+3)}{25-3}\\=\frac{2\times 28}{22}\\=\frac{28}{11}--(1)$

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

$=\frac{-4\times 5\times \sqrt{3}}{(5^{2}-\left(\sqrt{3}\right)^{2}}$

$=-\frac{20\sqrt{3}}{(25-3)}\\=-\frac{20\sqrt{3}}{22}\\=-\frac{10\sqrt{3}}{11}--(2)$

$Now,\\x^{2}-y^{2}\\=(x+y)(x-y)\\=\frac{28}{11}\times \frac{-10\sqrt{3}}{11}\\=\frac{-280\sqrt{3}}{121}$

•••♪