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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

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

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

Thanks sir, i appreciate.

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

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

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

Commented by mr W last updated on 21/Jun/20

Commented by mr W last updated on 21/Jun/20

if there are three real roots, P and  Q must lie above and under the   x−axis.

$${if}\:{there}\:{are}\:{three}\:{real}\:{roots},\:{P}\:{and} \\ $$$${Q}\:{must}\:{lie}\:{above}\:{and}\:{under}\:{the}\: \\ $$$${x}−{axis}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 21/Jun/20

let a and b be the roots of f(x) then  f(x)=(x−a)^2 (x−b)  f′(x)=2(x−a)(x−b)+(x−a)^2   it is clear that a is also the root of f′(x). proof

$$\mathrm{let}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}\:\mathrm{be}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{then} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} \left(\mathrm{x}−\mathrm{b}\right) \\ $$$$\boldsymbol{\mathrm{f}}'\left(\mathrm{x}\right)=\mathrm{2}\left(\mathrm{x}−\mathrm{a}\right)\left(\mathrm{x}−\mathrm{b}\right)+\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{clear}\:\mathrm{that}\:\mathrm{a}\:\mathrm{is}\:\mathrm{also}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:\boldsymbol{\mathrm{f}}'\left(\mathrm{x}\right).\:\boldsymbol{\mathrm{proof}} \\ $$

Commented by mr W last updated on 21/Jun/20

Commented by mr W last updated on 21/Jun/20

if there is a double root, one from  P and Q must lie on the x−axis.

$${if}\:{there}\:{is}\:{a}\:{double}\:{root},\:{one}\:{from} \\ $$$${P}\:{and}\:{Q}\:{must}\:{lie}\:{on}\:{the}\:{x}−{axis}. \\ $$

Commented by mr W last updated on 21/Jun/20

Commented by mr W last updated on 21/Jun/20

if there is only one real root, then  P and Q don′t exist or both of them  must lie above or under the x−axis.

$${if}\:{there}\:{is}\:{only}\:{one}\:{real}\:{root},\:{then} \\ $$$${P}\:{and}\:{Q}\:{don}'{t}\:{exist}\:{or}\:{both}\:{of}\:{them} \\ $$$${must}\:{lie}\:{above}\:{or}\:{under}\:{the}\:{x}−{axis}. \\ $$

Commented by mr W last updated on 21/Jun/20

a double root means: the curve f(x)  tangents the x−axis.

$${a}\:{double}\:{root}\:{means}:\:{the}\:{curve}\:{f}\left({x}\right) \\ $$$${tangents}\:{the}\:{x}−{axis}. \\ $$

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