Question-99430

Question Number 99430 by I want to learn more last updated on 20/Jun/20

Answered by mr W last updated on 20/Jun/20

f(x)=x^3 +(1/6)x^2 −((44)/9)x−((40)/9)=0 f ′(x)=3x^2 +(1/3)x−((44)/9)=0 ⇒27x^2 +3x−44=0 ⇒x=−(4/3), ((11)/9) one of them is a double root of f(x)=0. after checking we know it is x=−(4/3). say the other root is β, then x^3 +(1/6)x^2 −((44)/9)x−((40)/9)=(x+(4/3))^2 (x−β) with x=0 we get −((40)/9)=((4/3))^2 (−β) ⇒β=(5/2) ⇒the roots are x=−(4/3),−(4/3) and (5/2)

$${f}\left({x}\right)={x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{2}} −\frac{\mathrm{44}}{\mathrm{9}}{x}−\frac{\mathrm{40}}{\mathrm{9}}=\mathrm{0} \\ $$$${f}\:'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{x}−\frac{\mathrm{44}}{\mathrm{9}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{27}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{44}=\mathrm{0} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{4}}{\mathrm{3}},\:\frac{\mathrm{11}}{\mathrm{9}} \\ $$$${one}\:{of}\:{them}\:{is}\:{a}\:{double}\:{root}\:{of}\:{f}\left({x}\right)=\mathrm{0}. \\ $$$${after}\:{checking}\:{we}\:{know}\:{it}\:{is}\:{x}=−\frac{\mathrm{4}}{\mathrm{3}}. \\ $$$${say}\:{the}\:{other}\:{root}\:{is}\:\beta,\:{then} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{2}} −\frac{\mathrm{44}}{\mathrm{9}}{x}−\frac{\mathrm{40}}{\mathrm{9}}=\left({x}+\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} \left({x}−\beta\right) \\ $$$${with}\:{x}=\mathrm{0}\:{we}\:{get} \\ $$$$−\frac{\mathrm{40}}{\mathrm{9}}=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} \left(−\beta\right) \\ $$$$\Rightarrow\beta=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{the}\:{roots}\:{are}\:{x}=−\frac{\mathrm{4}}{\mathrm{3}},−\frac{\mathrm{4}}{\mathrm{3}}\:{and}\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by floor(10²Eta[1]) last updated on 21/Jun/20

$$ \\ $$$$ \\ $$$${ok}\:{but}\:{what}'{s}\:{the}\:{proof} \\ $$

Commented by I want to learn more last updated on 20/Jun/20

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$

Commented by floor(10²Eta[1]) last updated on 21/Jun/20

can you explain me why doing f′(x)=0 you know that one of the roots are repeated∫

$$ \\ $$$$ \\ $$$${can}\:{you}\:{explain}\:{me}\:{why}\:{doing}\:{f}'\left({x}\right)=\mathrm{0} \\ $$$${you}\:{know}\:{that}\:{one}\:{of}\:{the}\:{roots}\:{are}\:{repeated}\int \\ $$

Commented by 1549442205 last updated on 21/Jun/20

we have the following property: if the equation f(x)=0 has one double root then that root is a root of the equation f ′(x)=0

$$\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{following}\:\mathrm{property}: \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0}\:\mathrm{has}\:\mathrm{one}\:\mathrm{double}\:\mathrm{root}\:\mathrm{then}\: \\ $$$$\mathrm{that}\:\mathrm{root}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{0} \\ $$