Question Number 13438 by Nayon last updated on 20/May/17
$$\mathrm{7}{x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{9}=\mathrm{0} \\ $$$${How}\:{many}\:{roots}\:{of}\:{this}\:{equation} \\ $$$${are}\:{Negative}? \\ $$$$ \\ $$
Answered by mrW1 last updated on 20/May/17
![7x^5 −4x^4 +9x^3 +12x^2 +5x−9=0 7x^5 −9x^3 +5x−4x^4 +12x^2 −9=0 x(7x^4 −9x^2 +5)=(4x^4 −12x^2 +9) 7x(x^4 −(9/7)x^2 +(5/7))=4(x^4 −3x^2 +(9/4)) 7x[x^4 −2×(9/(14))x^2 +((9/(14)))^2 +(5/7)−((9/(14)))^2 ]=4[x^4 −2×(3/2)x^2 +((3/2))^2 ] 7x[(x^2 −(9/(14)))^2 +((59)/(196))]=4(x^2 −(3/2))^2 x=((4(x^2 −(3/2))^2 )/(7[(x^2 −(9/(14)))^2 +((59)/(196))]))=((>0)/(>0))>0 ⇒there is no negative root!](Q13448.png)
$$\mathrm{7}{x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{7}{x}^{\mathrm{5}} −\mathrm{9}{x}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{4}{x}^{\mathrm{4}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{9}=\mathrm{0} \\ $$$${x}\left(\mathrm{7}{x}^{\mathrm{4}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{5}\right)=\left(\mathrm{4}{x}^{\mathrm{4}} −\mathrm{12}{x}^{\mathrm{2}} +\mathrm{9}\right) \\ $$$$\mathrm{7}{x}\left({x}^{\mathrm{4}} −\frac{\mathrm{9}}{\mathrm{7}}{x}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{7}}\right)=\mathrm{4}\left({x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{4}}\right) \\ $$$$\mathrm{7}{x}\left[{x}^{\mathrm{4}} −\mathrm{2}×\frac{\mathrm{9}}{\mathrm{14}}{x}^{\mathrm{2}} +\left(\frac{\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{7}}−\left(\frac{\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} \right]=\mathrm{4}\left[{x}^{\mathrm{4}} −\mathrm{2}×\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \right] \\ $$$$\mathrm{7}{x}\left[\left({x}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} +\frac{\mathrm{59}}{\mathrm{196}}\right]=\mathrm{4}\left({x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{4}\left({x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{7}\left[\left({x}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} +\frac{\mathrm{59}}{\mathrm{196}}\right]}=\frac{>\mathrm{0}}{>\mathrm{0}}>\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{there}\:{is}\:{no}\:{negative}\:{root}! \\ $$
Commented by ajfour last updated on 20/May/17
$${awesome}\:! \\ $$