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Question Number 190717 by sciencestudentW last updated on 09/Apr/23

if x^2 +(m−2)x+2m=0 and (x_1 −1)(x_2 −1)=1 then find the value of m=?

$${if}\:\:{x}^{\mathrm{2}} +\left({m}−\mathrm{2}\right){x}+\mathrm{2}{m}=\mathrm{0}\:\:{and} \\ $$$$\left({x}_{\mathrm{1}} −\mathrm{1}\right)\left({x}_{\mathrm{2}} −\mathrm{1}\right)=\mathrm{1}\:\:{then}\:{find}\:{the}\:{value} \\ $$$${of}\:\:\:{m}=? \\ $$

Answered by mahdipoor last updated on 10/Apr/23

(x_1 −1)(x_2 −1)=x_1 x_2 +1−(x_1 +x_2 )= ((2m)/1)+1−((−(m−2))/1)=1⇒3m=2⇒m=2/3 note , ax^2 +bx+c=0 ⇒ x_1 ,x_2 =((−b±(√(b^2 −4ac)))/(2a)) ⇒x_1 +x_2 =((−b)/a) x_1 x_2 =(c/a)

$$\left({x}_{\mathrm{1}} −\mathrm{1}\right)\left({x}_{\mathrm{2}} −\mathrm{1}\right)={x}_{\mathrm{1}} {x}_{\mathrm{2}} +\mathrm{1}−\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)= \\ $$$$\frac{\mathrm{2}{m}}{\mathrm{1}}+\mathrm{1}−\frac{−\left({m}−\mathrm{2}\right)}{\mathrm{1}}=\mathrm{1}\Rightarrow\mathrm{3}{m}=\mathrm{2}\Rightarrow{m}=\mathrm{2}/\mathrm{3} \\ $$$${note}\:,\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} ,{x}_{\mathrm{2}} =\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\frac{−{b}}{{a}}\:\:\:\:\:\:\:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} =\frac{{c}}{{a}} \\ $$

Answered by cortano12 last updated on 10/Apr/23

⇒x^2 +(m−2)x+2m=0 { (x_1 ),(x_2 ) :} ⇒(x+1)^2 +(m−2)(x+1)+2m=0 { ((x_1 −1)),((x_2 −1)) :} ⇒x^2 +2x+1+(m−2)x+(m−2)+2m=0 ⇒x^2 +mx+3m−1=0 { ((x_1 −1)),((x_2 −1)) :} ∴ (x_1 −1)(x_2 −1)= 3m−1 ⇒ 1 = 3m−1 ; m=(2/3)

$$\:\Rightarrow\mathrm{x}^{\mathrm{2}} +\left(\mathrm{m}−\mathrm{2}\right)\mathrm{x}+\mathrm{2m}=\mathrm{0\begin{cases}{\mathrm{x}_{\mathrm{1}} }\\{\mathrm{x}_{\mathrm{2}} }\end{cases}} \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{m}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{2m}=\mathrm{0\begin{cases}{\mathrm{x}_{\mathrm{1}} −\mathrm{1}}\\{\mathrm{x}_{\mathrm{2}} −\mathrm{1}}\end{cases}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}+\left(\mathrm{m}−\mathrm{2}\right)\mathrm{x}+\left(\mathrm{m}−\mathrm{2}\right)+\mathrm{2m}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{mx}+\mathrm{3m}−\mathrm{1}=\mathrm{0\begin{cases}{\mathrm{x}_{\mathrm{1}} −\mathrm{1}}\\{\mathrm{x}_{\mathrm{2}} −\mathrm{1}}\end{cases}} \\ $$$$\:\therefore\:\left(\mathrm{x}_{\mathrm{1}} −\mathrm{1}\right)\left(\mathrm{x}_{\mathrm{2}} −\mathrm{1}\right)=\:\mathrm{3m}−\mathrm{1} \\ $$$$\:\Rightarrow\:\mathrm{1}\:=\:\mathrm{3m}−\mathrm{1}\:;\:\mathrm{m}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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