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

Commented by tanmay last updated on 18/Jun/19

question is not clear...

$${question}\:{is}\:{not}\:{clear}... \\ $$

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}

$$\alpha{x}^{\mathrm{2}} +\beta{x}+\gamma=\mathrm{0} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} =\frac{−\beta−\sqrt{\beta^{\mathrm{2}} −\mathrm{4}\alpha\gamma}}{\mathrm{2}\alpha},\:\:{x}_{\mathrm{2}} =\frac{−\beta+\sqrt{\beta^{\mathrm{2}} −\mathrm{4}\alpha\gamma}}{\mathrm{2}\alpha} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\frac{\beta}{\alpha} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\frac{\mathrm{2}{a}+\mathrm{3}}{{a}+\mathrm{1}}={y}\:\Rightarrow \\ $$$$\Rightarrow\:{a}=−\frac{{y}+\mathrm{3}}{{y}+\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)\in\mathbb{R}\backslash\left\{−\mathrm{2}\right\} \\ $$

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

$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Sir}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{taken}}\:\boldsymbol{\mathrm{both}}\:\boldsymbol{\mathrm{cases}}\:\boldsymbol{\mathrm{x}}_{\mathrm{1}} \:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} \:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{negative}}? \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{As}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{negative}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{other}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{positivet}}? \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Ans}}\:\boldsymbol{\mathrm{is}}\:\mathrm{2} \\ $$

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=−((0+3)/(0+2))=−(3/2)

$$\mathrm{I}\:\mathrm{corrected}\:\mathrm{a}\:\mathrm{typo} \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{equals} \\ $$$$\mathrm{zero}: \\ $$$${a}=−\frac{\mathrm{0}+\mathrm{3}}{\mathrm{0}+\mathrm{2}}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by azizullah last updated on 19/Jun/19

Ok my dear sir.

$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Ok}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{dear}}\:\boldsymbol{\mathrm{sir}}. \\ $$$$\:\:\:\:\:\:\:\: \\ $$

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