Question

a = 1/(3 - sqrt(11)) Find a ^ 2 - b ^ 2​

Answer 1

Solution -

a = 1/3-√11

It is given that b = 1/a = 3 + √11

Let us rationalise a first.

a = 1/3-√11 = (3+√11)/(3-√11)(3+√11)

> a = (3+√11)/-2

Now we need to find the value of a² - b²

a = ( 3+√11)/-2

a² > [ 3 + √11 ]²/4

> [ 9 + 11 + 6√11]/4

> [ 20 + 6√11]/4

> 5 + 3/2 √11

b² = ( 3 + √11 )²

> 9 + 11 + 6√11

> 20 + 6√11

a² - b²

> 5 + 3/2√11 - 20 - 6√11

> -15 - 4√11 .

This is the required answer .

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Answer 2

Answer:

Step-by-step explanation:

$a = \frac{1}{3-\sqrt{11}} \\b=\frac{1}{a}=\frac{1}{\frac{1}{3-\sqrt{11}}}=3-\sqrt{11}\\a^2-b^2 = (\frac{1}{3-\sqrt{11}})^2 - (3-\sqrt{11})^2\\= \frac{1}{(3-\sqrt{11})^2} - (3-\sqrt{11})^2\\=\frac{1- (3-\sqrt{11})^4}{(3-\sqrt{11})^2}\\=\frac{1 - ((3^4)-4(3)^3(\sqrt{11})+6(3)^2(\sqrt{11})^2-4(3)(\sqrt{11})^3+(\sqrt{11})^4)}{(3)^2 -2(3)(\sqrt{11})+(\sqrt{11})^2}\\=\frac{1 - (3^4)+4(3)^3(\sqrt{11})-6(3)^2(\sqrt{11})^2+4(3)(\sqrt{11})^3-(\sqrt{11})^4}{9 -6\sqrt{11}+11}$

$=\frac{1-81+108\sqrt{11}-(54 \times 11)+(12 \times 11)\sqrt{11}-121}{20-6\sqrt{11}}\\=\frac{-795+240\sqrt{11}}{20-6\sqrt{11}}$