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

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))

Commented bymathocean1 last updated on 19/Jul/20

Thanks.

Commented byabdomathmax last updated on 20/Jul/20

you are 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))★

Commented bymathocean1 last updated on 20/Jul/20

thanks