akiyamamayumi1 akiyamamayumi1
  • 18-08-2018
  • Mathematics
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p^2q^2x^2-(4q^4-3p^4)x-12p^2q^2=0 using quadratic formula

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gmany
gmany gmany
  • 19-08-2018

[tex] p^2q^2x^2-(4q^4-3p^4)x-12p^2q^2=0\\\\a=p^2q^2,\ b=-(4q^4-3p^4);\ c=-12p^2q^2\\\\b^2-4ac=[-(4q^4-3p^4)]^2-4(p^2q^2)(-12p^2q^2)\\\\=(4q^4)^2-2(4q^4)(3p^4)+(3p^4)^2+48p^4q^4\\\\=16q^8-24q^4p^4+9p^8+48p^4q^4=16q^8+24p^4q^4+9p^8 [/tex]

[tex] x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_1=\dfrac{4q^4-3p^4-\sqrt{16q^8+24p^4q^4+9p^8}}{2p^2q^2}\\\\x_2=\dfrac{4q^4-3p^4+\sqrt{16q^8+24p^4q^4+9p^8}}{2p^2q^2}\\\\\text{The domain}\\D:p\neq0\ \wedge\ q\neq0 [/tex]

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