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

$i f x^{2} + (m - 2) x + 2 m = 0 a n d$ $(x_{1} - 1) (x_{2} - 1) = 1 t h e n f i n d t h e v a l u e$ $o f m = ?$
	Answered by mahdipoor last updated on 10/Apr/23

	$(x_{1} - 1) (x_{2} - 1) = x_{1} x_{2} + 1 - (x_{1} + x_{2}) =$ $\frac{2 m}{1} + 1 - \frac{- (m - 2)}{1} = 1 \Rightarrow 3 m = 2 \Rightarrow m = 2 / 3$ $n o t e, {a x}^{2} + b x + c = 0 \Rightarrow$ $x_{1}, x_{2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $\Rightarrow x_{1} + x_{2} = \frac{- b}{a} x_{1} x_{2} = \frac{c}{a}$
	Answered by cortano12 last updated on 10/Apr/23
	$⇒x^2 +(m−2)x+2m=0 { (x_1 ),(x_2 ) :} ⇒(x+1)^2 +(m−2)(x+1)+2m=0 { ((x_1 −1)),((x_2 −1)) :} ⇒x^2 +2x+1+(m−2)x+(m−2)+2m=0 ⇒x^2 +mx+3m−1=0 { ((x_1 −1)),((x_2 −1)) :} ∴ (x_1 −1)(x_2 −1)= 3m−1 ⇒ 1 = 3m−1 ; m=(2/3)$
	$\Rightarrow x^{2} + (m - 2) x + 2 m = 0 {\begin{cases} x_{1} \\ x_{2} \end{cases}$ $\Rightarrow {(x + 1)}^{2} + (m - 2) (x + 1) + 2 m = 0 {\begin{cases} x_{1} - 1 \\ x_{2} - 1 \end{cases}$ $\Rightarrow x^{2} + 2 x + 1 + (m - 2) x + (m - 2) + 2 m = 0$ $\Rightarrow x^{2} + mx + 3 m - 1 = 0 {\begin{cases} x_{1} - 1 \\ x_{2} - 1 \end{cases}$ $∴ (x_{1} - 1) (x_{2} - 1) = 3 m - 1$ $\Rightarrow 1 = 3 m - 1; m = \frac{2}{3}$
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