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  • Brainly User
2015-01-06T21:46:05+05:30
Write 28-16\sqrt{3} as (1)^{4}-4(1)^{3}(\sqrt{3})+6(1)^{2}(\sqrt{3})^{2}-4(1)(\sqrt{3})^{3}+(\sqrt{3})^{4} .
This is expansion of (1-\sqrt{3})^{4}.
Hence, answer is (1-\sqrt{3}) .
3 4 3
can this method be adopted by every one to solve any problem of this kind ?
I guess no. But questions like these are made by taking some irrational number and then raising it to some power and we are asked to find that irrational number.
there are at least two roots. root 3 - 1 and 1 - root 3.
2015-01-06T22:15:59+05:30

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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.
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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

1 5 1
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