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if-x-2-m-2-x-2m-0-and-x-1-1-x-2-1-1-then-find-the-value-of-m-

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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=?
ifx2+(m−2)x+2m=0and(x1−1)(x2−1)=1thenfindthevalueofm=?
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)
(x1−1)(x2−1)=x1x2+1−(x1+x2)=2m1+1−−(m−2)1=1⇒3m=2⇒m=2/3note,ax2+bx+c=0⇒x1,x2=−b±b2−4ac2a⇒x1+x2=−bax1x2=ca
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)
⇒x2+(m−2)x+2m=0{x1x2⇒(x+1)2+(m−2)(x+1)+2m=0{x1−1x2−1⇒x2+2x+1+(m−2)x+(m−2)+2m=0⇒x2+mx+3m−1=0{x1−1x2−1∴(x1−1)(x2−1)=3m−1⇒1=3m−1;m=23

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