(4+√15)^(3/2)-(4-√15)^(3/2)=K√6

Question

Value of k is required in this plz give ans

Answer 1

You can use a calculator and determine that k = 9.

However, solving algebraically, Square both sides. This will allow 3/2 powers, which we can't work with, become cubics.

Note that (x-y) ^ 2 = x^2 - 2xy + y^2

Thus, we have

(4+sqrt(15))^3 -2 (4+sqrt(15))^3/2 (4 - sqrt(15))^(3/2) + (4-sqrt(15))^3 = 6 k^2

Note that, while we still have (3/2) powers, (4-sqrt(15))(4 + sqrt(15)) = 1, so this is not an issue.

We have

(4+sqrt(15))^3 - 2 + (4-sqrt(15))^3 = 6 k^2

Then, note that (x+y)^3 = x^3 + 3 x^2 y + 3 xy^2 + y^3.

Lastly, note that, on the left, letting x = 4 and y = sqrt(15) and - sqrt(15), the terms with sqrt(15) to an odd power will cancel out due to the plus and minus.

Thus, we are left with 2 * (4^3 + 3 * 4 * 15) - 2 = 6 k^2

2 * (64 + 180) - 2 = 6 k^2

486 = 6 k^2

81 = k^2

k = +- 9

Squaring introduces extraneous roots. Clearly, (4+sqrt(15))^(3/2) > 1 and (4-sqrt(15))^(3/2) < 1, so the left side is positive, and k will be positive.

k = 9

Answer 2

I will not solve this for you, but I will tell you how to do it.

Square both sides of the awful looking equation.

(4+√15)³ - 2[(4+√15)(4-√15)]^3/2 + (4-√15)³ = 6K²

Expand the terms carefully. When you expand and add the 2 cubic terms, the terms with √15 will add out; the middle term becomes a whole number.

Now you will have a very simple equation to solve for K

Of course, you could just do the arithmetic on your calculator directly...and you would get messed up by round off error!

Answer 3

(4+√15)^(3/2)-(4-√15)^(3/2)=K√6

The exponent is the same in both terms, so

(4 + √15 - 4 + √15)63/2 = K√6

(2√15)^3/2 = K√6

square 2√15 = 60

cube root 60 = K√6

3.914 = K(2.45)