How do you solve this equation:   x^2= 21-√(x^2-9)

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`x^2 = 21-sqrt(x^2-9)`

 

Let `x^2-9 = k^2`

Then `sqrt(x^2-9) = +-k` and `x^2 = k^2+9`

 

`x^2 = 21-sqrt(x^2-9)`

`k^2+9 = 21-(+-k)`

 

First let us consider `sqrt(x^2-9) = +k` .

 

`k^2+9 = 21-k`

`k^2+k-12 = 0`

`k^2+4k-3k-12 = 0`

`k(k+4)-3(k+4) = 0`

`(k+4)(k-3) = 0`

 

k = -4 or k = 3

 

When k = -4

`x^2-9 = k^2`

`x^2-9 = (-4)^2`

`x^2-9 = 16`

  ` x^2 = 25`

      ` x = +-5`

 

When k = 3

`x^2-9 = k^2`

`x^2-9 = 3^2`

  ` x^2 = 18`

       `x = +-sqrt18`

 

 

let us consider `sqrt(x^2-9) = -k` .

 

`k^2+9 = 21-(-k)`

`k^2-k-12 = 0`

`k^2-4k+3k-12 = 0`

`k(k-4)+3(k-4) = 0`

`(k-4)(k+3) = 0`

 

k = 4 or k = -3

 

When k = 4

`x^2-9 = k^2`

`x^2-9 = (4)^2`

`x^2-9 = 16`

  ` x^2 = 25`

       `x = +-5`

 

When k = -3

`x^2-9 = k^2`

`x^2-9 = (-3)^2`

   `x^2 = 18`

       `x = +-sqrt18`

 

 

If you consider both + and - answers for `sqrt(x^2-9)` the answers are;

`x = -5,-sqrt18,5,sqrt18`

 

If you consider `sqrt(x^2-9)` >0 always ` ` the answer is;

x = sqrt18

 

 

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