Menu Close

Question-83554




Question Number 83554 by Power last updated on 03/Mar/20
Answered by MJS last updated on 03/Mar/20
x^4 +4x^3 −8x+((17)/5)=0  let x=t−1  t^4 −6t^2 +((42)/5)=0  t^2 =3±((√(15))/5)  t=±(√(3±((√(15))/5)))  x=−1±(√(3±((√(15))/5)))
$${x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} −\mathrm{8}{x}+\frac{\mathrm{17}}{\mathrm{5}}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}={t}−\mathrm{1} \\ $$$${t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\frac{\mathrm{42}}{\mathrm{5}}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =\mathrm{3}\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{5}} \\ $$$${t}=\pm\sqrt{\mathrm{3}\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{5}}} \\ $$$${x}=−\mathrm{1}\pm\sqrt{\mathrm{3}\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{5}}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *