# Find the fourth root of 28-16 √3 without using log tables. But use short method. I need it urgently.

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by anudeepd2

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by anudeepd2

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(x - y√3)^4 = 28 - 16 √3

x⁴ - 4 x³ y√3 + 6 x² * 3 y² - 4 x y³ 3√3 + 9 y⁴ = 28 - 16 √3

x⁴ + 18 x²y² + 9 y⁴ = 28 -- equation 1

x² y + 3 x y² = 4 -- equation 2

Solve them, if you can. Or follow the method below. Perhaps there is a simpler method too.

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I do that in two smaller steps . if i cannot do more complex way.

(x - y√3)² = 28 - 16 √3

Hence, x² + 3 y² = 28 and 2 x y = 16 => y = 8 / x

x² + 3 * (8/x)² = 28

x⁴ - 28 x² + 192 = 0

x² = [ 28 + - √(28² - 4*192) ] / 2 = 16 or 12 Hence x = +4 or -4 or √12 or -√12

y = +2, - 2, +4/√3 , - 4/√3.

You can factorise above as (x² - 16) (x² -12 ) = 0

Now square root of 28 - 16 √3 = 4 - 2 √3 . We can take negative values also. But then their square roots will be imaginary. You can also that do if required.

============================

Now find the square root of 4 - 2 √3 like above.

(x - y√3)² = 4 - 2 √3

solving like above

x² = 3 or 1

so x = -1, + 1, + √3 or -√3.

Hence, answer is 1- √3 or √3 - 1

x⁴ - 4 x³ y√3 + 6 x² * 3 y² - 4 x y³ 3√3 + 9 y⁴ = 28 - 16 √3

x⁴ + 18 x²y² + 9 y⁴ = 28 -- equation 1

x² y + 3 x y² = 4 -- equation 2

Solve them, if you can. Or follow the method below. Perhaps there is a simpler method too.

====================================

I do that in two smaller steps . if i cannot do more complex way.

(x - y√3)² = 28 - 16 √3

Hence, x² + 3 y² = 28 and 2 x y = 16 => y = 8 / x

x² + 3 * (8/x)² = 28

x⁴ - 28 x² + 192 = 0

x² = [ 28 + - √(28² - 4*192) ] / 2 = 16 or 12 Hence x = +4 or -4 or √12 or -√12

y = +2, - 2, +4/√3 , - 4/√3.

You can factorise above as (x² - 16) (x² -12 ) = 0

Now square root of 28 - 16 √3 = 4 - 2 √3 . We can take negative values also. But then their square roots will be imaginary. You can also that do if required.

============================

Now find the square root of 4 - 2 √3 like above.

(x - y√3)² = 4 - 2 √3

solving like above

x² = 3 or 1

so x = -1, + 1, + √3 or -√3.

Hence, answer is 1- √3 or √3 - 1