# How can √((x^4)+(3x^2))-(x^2) become 3/2?

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You need to complete the square `x^4 + 3x^2` using the formula `(a+b)^2 = a^2 + 2ab + b^2` .

`x^4 + 3x^2 = a^2 + 2ab => {(x^4 = a^2 => x^2=a),(3x^2 = 2x^2b => b = 3/2):}`

Completing the square yields:

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

Hence, substituting `sqrt(x^4 + 3x^2) - x^2 = sqrt((x^2 + 3/2)^2 - 9/4) - x^2 != 3/2`

**Hence, completing the square yields `(x^2 + 3/2)^2 - 9/4` , thus `sqrt(x^4 + 3x^2) - x^2 = sqrt((x^2 + 3/2)^2 - 9/4) - x^2 != 3/2.` **

take **x^2** on RHS

SQUARE on both sides

so **LHS=(x^4 + 3x^2)**

and RHS=**( x^4 + 3x^2 +9/4)**

on solving it gives **9/4 = 0 ** which is impossible.

**SO WE CAN SAY THAT THIS EQUATION HAVE NO SOLUTIONS**