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Question Number 140974 by EnterUsername last updated on 14/May/21

If the equations x^2 −3x+a=0 and x^2 +ax−3=0 have a common root, then a possible value of a is (A) 3 (B) 1 (C) −2 (D) 2

$$\mathrm{If}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+{a}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} +{a}\mathrm{x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root},\:\mathrm{then}\:\mathrm{a}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{2} \\ $$

Commented by mohammad17 last updated on 14/May/21

x^2 +ax−3=0→x+a−(3/x)=0→(1) x^2 −3x+a=0→(2) from (2)−(1) x^2 −4x+(3/x)=0⇒x^3 −4x^2 +3=0 x=1 , x=((3−(√(21)))/2) , x=((3+(√(21)))/2) since : x=1⇒a=2 since:x=((3+(√(21)))/2)⇒a=−3 cancel since x=((3−(√(21)))/2)⇒a=−3 cance ⇒a=2

$$ \\ $$$${x}^{\mathrm{2}} +{ax}−\mathrm{3}=\mathrm{0}\rightarrow{x}+\mathrm{a}−\frac{\mathrm{3}}{\mathrm{x}}=\mathrm{0}\rightarrow\left(\mathrm{1}\right) \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+{a}=\mathrm{0}\rightarrow\left(\mathrm{2}\right) \\ $$$$ \\ $$$${from}\:\left(\mathrm{2}\right)−\left(\mathrm{1}\right) \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\frac{\mathrm{3}}{{x}}=\mathrm{0}\Rightarrow{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}=\mathrm{0} \\ $$$$ \\ $$$${x}=\mathrm{1}\:\:\:,\:\:{x}=\frac{\mathrm{3}−\sqrt{\mathrm{21}}}{\mathrm{2}}\:\:,\:{x}=\frac{\mathrm{3}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$ \\ $$$${since}\::\:{x}=\mathrm{1}\Rightarrow{a}=\mathrm{2} \\ $$$$ \\ $$$${since}:{x}=\frac{\mathrm{3}+\sqrt{\mathrm{21}}}{\mathrm{2}}\Rightarrow{a}=−\mathrm{3}\:\:{cancel} \\ $$$$ \\ $$$${since}\:{x}=\frac{\mathrm{3}−\sqrt{\mathrm{21}}}{\mathrm{2}}\Rightarrow{a}=−\mathrm{3}\:{cance} \\ $$$$ \\ $$$$\Rightarrow{a}=\mathrm{2} \\ $$

Commented by EnterUsername last updated on 14/May/21

Thanks you!

$$\mathrm{Thanks}\:\mathrm{you}! \\ $$

Commented by mohammad17 last updated on 14/May/21

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Answered by henderson last updated on 14/May/21

the answer is a = 2.

$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:{a}\:=\:\mathrm{2}. \\ $$

Answered by physicstutes last updated on 14/May/21

I will say from experience a = 2 , but if you need a proper working step, you could say let the first equation have roots α and β and let the second equation have root β and ϑ notice that β is the common root. α + β = 3.....(i) and αβ = a .....(ii) β + ϑ = −a .....(ii) and βϑ = −3.....(iv) from (iv) ϑ =−(3/β) ⇒ β−(3/β)=−a also α = (a/β) ⇒ (a/β)+β = 3 and (3/β)−β = a a + β^2 = 3β and 3−β^2 = aβ add them and get: 3 + a = β(3+a) ⇒ β = 1 now 1−(3/1)=−a ⇒ a = 2

$$\mathrm{I}\:\mathrm{will}\:\mathrm{say}\:\mathrm{from}\:\mathrm{experience}\:\:{a}\:=\:\mathrm{2}\:,\:\mathrm{but}\:\mathrm{if}\:\mathrm{you}\:\mathrm{need}\:\mathrm{a}\:\mathrm{proper} \\ $$$$\mathrm{working}\:\mathrm{step},\:\mathrm{you}\:\mathrm{could}\:\mathrm{say}\: \\ $$$$\mathrm{let}\:\mathrm{the}\:\mathrm{first}\:\mathrm{equation}\:\mathrm{have}\:\mathrm{roots}\:\alpha\:\mathrm{and}\:\beta\: \\ $$$$\mathrm{and}\:\mathrm{let}\:\mathrm{the}\:\mathrm{second}\:\mathrm{equation}\:\mathrm{have}\:\mathrm{root}\:\beta\:\mathrm{and}\:\vartheta\: \\ $$$$\mathrm{notice}\:\mathrm{that}\:\beta\:\mathrm{is}\:\mathrm{the}\:\mathrm{common}\:\mathrm{root}. \\ $$$$\alpha\:+\:\beta\:=\:\mathrm{3}.....\left({i}\right)\:\:\:\mathrm{and}\:\:\:\alpha\beta\:=\:{a}\:.....\left({ii}\right) \\ $$$$\beta\:+\:\vartheta\:=\:−{a}\:.....\left({ii}\right)\:\:\mathrm{and}\:\beta\vartheta\:=\:−\mathrm{3}.....\left({iv}\right) \\ $$$$\mathrm{from}\:\left({iv}\right)\:\:\vartheta\:=−\frac{\mathrm{3}}{\beta} \\ $$$$\:\Rightarrow\:\:\:\beta−\frac{\mathrm{3}}{\beta}=−{a}\:\: \\ $$$$\mathrm{also}\:\:\alpha\:=\:\frac{{a}}{\beta} \\ $$$$\Rightarrow\:\:\:\frac{{a}}{\beta}+\beta\:=\:\mathrm{3}\:\:\mathrm{and}\:\:\frac{\mathrm{3}}{\beta}−\beta\:=\:{a} \\ $$$${a}\:+\:\beta^{\mathrm{2}} \:=\:\mathrm{3}\beta \\ $$$$\mathrm{and}\:\:\:\mathrm{3}−\beta^{\mathrm{2}} \:=\:{a}\beta \\ $$$$\mathrm{add}\:\mathrm{them}\:\mathrm{and}\:\mathrm{get}:\:\:\mathrm{3}\:+\:{a}\:=\:\beta\left(\mathrm{3}+{a}\right)\:\:\Rightarrow\:\beta\:=\:\mathrm{1} \\ $$$$\mathrm{now}\:\:\:\mathrm{1}−\frac{\mathrm{3}}{\mathrm{1}}=−{a}\:\:\Rightarrow\:{a}\:=\:\mathrm{2}\: \\ $$

Commented by EnterUsername last updated on 14/May/21

Thanks for the idea!

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{idea}! \\ $$

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