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-

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

G i v e n :

(E) : (m + 1) x^{2} + (m - 2) x + 1 = 0

W e s u p p o s e t h a t i t h a s t w o r o o t s x_{1}

a n d x_{2} .

D e t e r m i n a t e m s u c h t h a t :

x_{1} = 1 + x_{2}

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

we have x_{1} + x_{2} = - \frac{m - 2}{m + 1} and x_{1} x_{2} = \frac{1}{m + 1} also

x_{1} = 1 + x_{2} \Rightarrow x_{1} - x_{2} = 1 \Rightarrow {2 x}_{1} = 1 - \frac{m - 2}{m + 1} = \frac{m + 1 - m + 2}{m + 1} = \frac{3}{m + 1}

{2 x}_{2} = - \frac{m - 2}{m + 1} - 1 = \frac{- m + 2 - m - 1}{m + 1} = \frac{- 2 m + 1}{m + 1}

{4 x}_{1} x_{2} = \frac{3}{m + 1} \times \frac{- 2 m + 1}{m + 1} = \frac{4}{m + 1} \Rightarrow \frac{3 (- 2 m + 1)}{{(m + 1)}^{2}} = \frac{4}{m + 1} \Rightarrow

\frac{- 6 m + 3}{m + 1} = 4 \Rightarrow - 6 m + 3 = 4 m + 4 \Rightarrow - 10 m = 1 \Rightarrow m = - \frac{1}{10}

Commented by mathocean1 last updated on 19/Jul/20

T h a n k s .

Commented by abdomathmax last updated on 20/Jul/20

you are welcome

Answered by bemath last updated on 20/Jul/20

x_{1} - x_{2} = 1 = \frac{- b + \sqrt{Δ}}{2 . a} - (\frac{- b - \sqrt{Δ}}{2 . a})

\Rightarrow \frac{\sqrt{Δ}}{a} = 1 \to Δ = a^{2}

m^{2} - 4 m + 4 - 4 (m + 1) = {(m + 1)}^{2}

m^{2} - 8 m = {(m + 1)}^{2}

m^{2} - 8 m = m^{2} + 2 m + 1 \Rightarrow m = - \frac{1}{10} ★

Commented by mathocean1 last updated on 20/Jul/20

t h a n k s