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Given-E-m-1-x-2-m-2-x-1-0-We-suppose-that-it-has-two-roots-x-1-and-x-2-Determinate-m-such-that-x-1-1-x-2-

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Question Number 104187 by mathocean1 last updated on 19/Jul/20
Given: (E): (m+1)x^2 +(m−2)x+1=0 We suppose that it has two roots x_1 and x_2 . Determinate m such that: x_1 =1+x_2
$${Given}: \\ $$$$\left({E}\right):\:\left({m}+\mathrm{1}\right){x}^{\mathrm{2}} +\left({m}−\mathrm{2}\right){x}+\mathrm{1}=\mathrm{0} \\ $$$${We}\:{suppose}\:{that}\:{it}\:{has}\:{two}\:{roots}\:{x}_{\mathrm{1}} \\ $$$${and}\:{x}_{\mathrm{2}} . \\ $$$${Determinate}\:{m}\:{such}\:{that}: \\ $$$${x}_{\mathrm{1}} =\mathrm{1}+{x}_{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 19/Jul/20
we have x_1 +x_2 =−((m−2)/(m+1)) and x_1 x_2 =(1/(m+1)) also x_1 =1+x_2 ⇒x_1 −x_2 =1 ⇒ 2x_1 =1−((m−2)/(m+1)) =((m+1−m+2)/(m+1)) =(3/(m+1)) 2x_2 =−((m−2)/(m+1))−1 =((−m+2−m−1)/(m+1)) =((−2m+1)/(m+1)) 4x_1 x_2 =(3/(m+1))×((−2m+1)/(m+1)) =(4/(m+1)) ⇒((3(−2m+1))/((m+1)^2 )) =(4/(m+1)) ⇒ ((−6m+3)/(m+1)) =4 ⇒−6m+3 =4m+4 ⇒−10m =1 ⇒m =−(1/(10))
$$\:\mathrm{we}\:\mathrm{have}\:\mathrm{x}_{\mathrm{1}} \:+\mathrm{x}_{\mathrm{2}} =−\frac{\mathrm{m}−\mathrm{2}}{\mathrm{m}+\mathrm{1}}\:\mathrm{and}\:\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{m}+\mathrm{1}}\:\:\mathrm{also} \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{1}+\mathrm{x}_{\mathrm{2}} \:\Rightarrow\mathrm{x}_{\mathrm{1}} −\mathrm{x}_{\mathrm{2}} =\mathrm{1}\:\Rightarrow\:\mathrm{2x}_{\mathrm{1}} =\mathrm{1}−\frac{\mathrm{m}−\mathrm{2}}{\mathrm{m}+\mathrm{1}}\:=\frac{\mathrm{m}+\mathrm{1}−\mathrm{m}+\mathrm{2}}{\mathrm{m}+\mathrm{1}}\:=\frac{\mathrm{3}}{\mathrm{m}+\mathrm{1}} \\ $$$$\mathrm{2x}_{\mathrm{2}} =−\frac{\mathrm{m}−\mathrm{2}}{\mathrm{m}+\mathrm{1}}−\mathrm{1}\:=\frac{−\mathrm{m}+\mathrm{2}−\mathrm{m}−\mathrm{1}}{\mathrm{m}+\mathrm{1}}\:=\frac{−\mathrm{2m}+\mathrm{1}}{\mathrm{m}+\mathrm{1}} \\ $$$$\mathrm{4x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{m}+\mathrm{1}}×\frac{−\mathrm{2m}+\mathrm{1}}{\mathrm{m}+\mathrm{1}}\:=\frac{\mathrm{4}}{\mathrm{m}+\mathrm{1}}\:\Rightarrow\frac{\mathrm{3}\left(−\mathrm{2m}+\mathrm{1}\right)}{\left(\mathrm{m}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{4}}{\mathrm{m}+\mathrm{1}}\:\Rightarrow \\ $$$$\frac{−\mathrm{6m}+\mathrm{3}}{\mathrm{m}+\mathrm{1}}\:=\mathrm{4}\:\Rightarrow−\mathrm{6m}+\mathrm{3}\:=\mathrm{4m}+\mathrm{4}\:\Rightarrow−\mathrm{10m}\:=\mathrm{1}\:\Rightarrow\mathrm{m}\:=−\frac{\mathrm{1}}{\mathrm{10}} \\ $$
Commented by mathocean1 last updated on 19/Jul/20
Thanks.
$${Thanks}. \\ $$
Commented by abdomathmax last updated on 20/Jul/20
you are welcome
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\: \\ $$
Answered by bemath last updated on 20/Jul/20
x_1 −x_2 = 1 = ((−b+(√Δ))/(2.a))−(((−b−(√Δ))/(2.a))) ⇒((√Δ)/a) = 1→Δ=a^2 m^2 −4m+4−4(m+1)=(m+1)^2 m^2 −8m= (m+1)^2 m^2 −8m=m^2 +2m+1⇒m=−(1/(10))★
$${x}_{\mathrm{1}} −{x}_{\mathrm{2}} \:=\:\mathrm{1}\:=\:\frac{−{b}+\sqrt{\Delta}}{\mathrm{2}.{a}}−\left(\frac{−{b}−\sqrt{\Delta}}{\mathrm{2}.{a}}\right) \\ $$$$\Rightarrow\frac{\sqrt{\Delta}}{{a}}\:=\:\mathrm{1}\rightarrow\Delta={a}^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} −\mathrm{4}{m}+\mathrm{4}−\mathrm{4}\left({m}+\mathrm{1}\right)=\left({m}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} \:−\mathrm{8}{m}=\:\left({m}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} −\mathrm{8}{m}={m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}\Rightarrow{m}=−\frac{\mathrm{1}}{\mathrm{10}}\bigstar \\ $$$$ \\ $$
Commented by mathocean1 last updated on 20/Jul/20
thanks
$${thanks} \\ $$

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