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.`
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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
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