solve this example of sum of squares: (x^2-9)^2+(x-3)^2(x^2+4x+3)^2=0

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

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`(x^2-9)^2+(x-3)^2(x^2+4x+3)^2=0`

 

`(x^2-9) `

 

 

 

 

`(x^2-9)^2`

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

`= (x-3)^2*(x+3)^2`

 

`x^2+4x+3`

`= x^2+3x+x+3`

`= x(x+3)+1(x+3)`

`= (x+3)(x+1)`

 

`(x^2+4x+3)^2`

`= (x+3)^2*(x+1)^2`

 

`(x^2-9)^2+(x-3)^2(x^2+4x+3)^2=0`

`(x-3)^2*(x+3)^2+(x-3)^2*(x+3)^2*(x+1)^2 = 0`

`(x-3)^2(x+3)^2(1+(x+1)^2) = 0`

 

`(x-3)^2 = 0`

        ` x = 3`

 

`(x+3)^2 = 0`

        ` x = -3`

 

`(1+(x+1)^2) = 0`

      `(x+1)^2 = -1`

Since `(x+1)^2>=0` this is not an answer.

 

So the answers are x = 3 and x = -3.

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