Caramtchll Caramtchll
  • 27-03-2014
  • Mathematics
Answered

find all solutions to 0= 2x^4+5x^3+38x^2+125x-300

Answer :

Аноним Аноним
  • 28-03-2014
[tex]2x^4+5x^3+38x^2+125x-300=0\\\\2x^4+5x^3-12x^2+50x^2+125x-300=0\\\\x^2(2x^2+5x-12)+25(2x^2+5x-12)+0\\\\(2x^2+5x-12)(x^2+25)=0\\\Updownarrow\\1^o\ 2x^2+5x-12=0\ \vee\ 2^o\ x^2+25=0\\\\\\1^o\ \Delta=5^2-4\cdot2\cdot(-12)=25+96=121\\\\\sqrt\Delta=\sqrt{121}=11\\\\x_1=\frac{-5-11}{2\cdot2}=\frac{-16}{4}=-4;\ x_2=\frac{-5+11}{2\cdot2}=\frac{6}{4}=\frac{3}{2}\\\\\\2^o\ x^2=-25-false\\\\\\Answer:x=-4\ or\ x=\frac{3}{2}[/tex]
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