binomial theorem

find the degree of the following expression:

(√2x2+1 + √2x2-1)6+ 26[√2x2+1 + √2x2-1 ]-6

6 Answers

11
virang1 Jhaveri ·

The degree is 6

1
Surbhi Shah ·

i need the solution not the answer
please explain

1
Kalyan Pilla ·

√2x2+1 = x[√2+(1/x2)]

Do similarly 4 the other term

This when raised to 6 gives x6[√{2+(1/x2)}+√{2-(1/x2)}]6

The [√{2+(1/x2)}+√{2-(1/x2)}]6 wala part part will giv U a constt. on expansion

Ise x6 se multiply karne pe U will get a term of power 6

The second term when expanded will have terms of power less than 6

So the degree of the expression is 6

1
Surbhi Shah ·

sorry i could nt get how did you get that the expansion of the second term will have powers less than 6

62
Lokesh Verma ·

[√2x2+1 + √2x2-1 ]

can be written as (\sqrt{2x^2+1}+\sqrt{2x^2-1})\frac{\sqrt{2x^2+1}-\sqrt{2x^2-1}}{\sqrt{2x^2+1}-\sqrt{2x^2-1}}

=\frac{2}{\sqrt{2x^2+1}-\sqrt{2x^2-1}}

Hence by replacing the second term the whole expression becomes

[√2x2+1 + √2x2-1 ]6+[√2x2+1 - √2x2-1 ]6

try with this in mind..

1
Surbhi Shah ·

thank you
what a lovely solution!!!

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