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Question Number 62265 by azizullah last updated on 18/Jun/19

Commented by tanmay last updated on 18/Jun/19

$q u e s t i o n i s n o t c l e a r . . .$
	Answered by MJS last updated on 19/Jun/19
	$αx^2 +βx+γ=0 ⇒ x_1 =((−β−(√(β^2 −4αγ)))/(2α)), x_2 =((−β+(√(β^2 −4αγ)))/(2α)) x_1 +x_2 =−(β/α) in our case x_1 +x_2 =−((2a+3)/(a+1))=y ⇒ ⇒ a=−((y+3)/(y+2)) ⇒ ⇒ (x_1 +x_2 )∈R\{−2}$
	$α x^{2} + β x + γ = 0$ $\Rightarrow x_{1} = \frac{- β - \sqrt{β^{2} - 4 α γ}}{2 α}, x_{2} = \frac{- β + \sqrt{β^{2} - 4 α γ}}{2 α}$ $x_{1} + x_{2} = - \frac{β}{α}$ $in our case x_{1} + x_{2} = - \frac{2 a + 3}{a + 1} = y \Rightarrow$ $\Rightarrow a = - \frac{y + 3}{y + 2} \Rightarrow$ $\Rightarrow (x_{1} + x_{2}) \in R ∖ {- 2}$
	Commented by azizullah last updated on 19/Jun/19

	$Sir you have taken both cases x_{1} and x_{2} are negative ?$ $As one is negative and other is positivet ?$ $Ans is 2$
	Commented by MJS last updated on 19/Jun/19

	$I corrected a typo$ $if you mean the sum of the the roots equals$ $zero :$ $a = - \frac{0 + 3}{0 + 2} = - \frac{3}{2}$
	Commented by azizullah last updated on 19/Jun/19

	$Ok my dear sir .$
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