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Question Number 104187 by mathocean1 last updated on 19/Jul/20

$$Given:$$$$\left(E\right):(m+1)x^2+(m-2)x+1=0$$$$Wesupposethatithastworoots{x}_1$$$$and{x}_2.$$$$Determinatemsuchthat:$$$$x_1=1+x_2$$

Answered by mathmax by abdo last updated on 19/Jul/20

$$\mathrm{we}\mathrm{have}{\mathrm{x}}_1+{\mathrm{x}}_2=-\frac{\mathrm{m}-2}{\mathrm{m}+1}\mathrm{and}{\mathrm{x}}_1{\mathrm{x}}_2=\frac{1}{\mathrm{m}+1}\mathrm{also}$$$${\mathrm{x}}_1=1+{\mathrm{x}}_2\Rightarrow {\mathrm{x}}_1-{\mathrm{x}}_2=1\Rightarrow {2\mathrm{x}}_1=1-\frac{\mathrm{m}-2}{\mathrm{m}+1}=\frac{\mathrm{m}+1-\mathrm{m}+2}{\mathrm{m}+1}=\frac{3}{\mathrm{m}+1}$$$${2\mathrm{x}}_2=-\frac{\mathrm{m}-2}{\mathrm{m}+1}-1=\frac{-\mathrm{m}+2-\mathrm{m}-1}{\mathrm{m}+1}=\frac{-2\mathrm{m}+1}{\mathrm{m}+1}$$$${4\mathrm{x}}_1{\mathrm{x}}_2=\frac{3}{\mathrm{m}+1}\times \frac{-2\mathrm{m}+1}{\mathrm{m}+1}=\frac{4}{\mathrm{m}+1}\Rightarrow \frac{3(-2\mathrm{m}+1)}{{(\mathrm{m}+1)}^2}=\frac{4}{\mathrm{m}+1}\Rightarrow$$$$\frac{-6\mathrm{m}+3}{\mathrm{m}+1}=4\Rightarrow -6\mathrm{m}+3=4\mathrm{m}+4\Rightarrow -10\mathrm{m}=1\Rightarrow \mathrm{m}=-\frac{1}{10}$$

Commented by mathocean1 last updated on 19/Jul/20

$$Thanks.$$

Commented by abdomathmax last updated on 20/Jul/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by bemath last updated on 20/Jul/20

$$x_1-x_2=1=\frac{-b+\sqrt{\mathrm{\Delta }}}{2.a}-\left(\frac{-b-\sqrt{\mathrm{\Delta }}}{2.a}\right)$$$$\Rightarrow \frac{\sqrt{\mathrm{\Delta }}}{a}=1\to \mathrm{\Delta }=a^2$$$$m^2-4m+4-4(m+1)={(m+1)}^2$$$$m^2-8m={(m+1)}^2$$$$m^2-8m=m^2+2m+1\Rightarrow m=-\frac{1}{10}★$$$$$$

Commented by mathocean1 last updated on 20/Jul/20

$$thanks$$

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