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Question-13141




Question Number 13141 by tawa tawa last updated on 15/May/17
Commented by RasheedSindhi last updated on 15/May/17
Mr tawa tawa, you are requested  to use bigger font.I felt much  difficulity to read this question.  (I made screen shot of your question   and then make it bigger with  my fingers!)
$$\mathrm{Mr}\:\mathrm{tawa}\:\mathrm{tawa},\:\mathrm{you}\:\mathrm{are}\:\mathrm{requested} \\ $$$$\mathrm{to}\:\mathrm{use}\:\mathrm{bigger}\:\mathrm{font}.\mathrm{I}\:\mathrm{felt}\:\mathrm{much} \\ $$$$\mathrm{difficulity}\:\mathrm{to}\:\mathrm{read}\:\mathrm{this}\:\mathrm{question}. \\ $$$$\left(\mathrm{I}\:\mathrm{made}\:\mathrm{screen}\:\mathrm{shot}\:\mathrm{of}\:\mathrm{your}\:\mathrm{question}\:\right. \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{make}\:\mathrm{it}\:\mathrm{bigger}\:\mathrm{with} \\ $$$$\left.\mathrm{my}\:\mathrm{fingers}!\right) \\ $$
Commented by tawa tawa last updated on 15/May/17
Sorry for the stress sir. God bless you sir.  But am a miss
$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{the}\:\mathrm{stress}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{But}\:\mathrm{am}\:\mathrm{a}\:\mathrm{miss} \\ $$
Commented by RasheedSindhi last updated on 15/May/17
Sorry miss tawa tawa!
$$\mathrm{Sorry}\:\mathrm{miss}\:\mathrm{tawa}\:\mathrm{tawa}! \\ $$
Answered by RasheedSindhi last updated on 15/May/17
(1)  x^2 −3x+2=0  x^2 −2x−x+2=0  x(x−2)−1(x−2)=0  (x−1)(x−2)=0  x−1=0 ∣ x−2=0  x=1 ∣  x=2  Solution Set:{1,2}={n∣n∈N∧n<3}  (2)  x^3 −1=0  (x−1)(x^2 +x+1)=0  x−1=0 ∣ x^2 +x+1=0  x=1 ∣ x=((−1±(√(1^2 −4(1)(1))))/(2(1)))  x=1, (1/2)(−1±i(√3))  Solution Set=B
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{x}\left(\mathrm{x}−\mathrm{2}\right)−\mathrm{1}\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{x}−\mathrm{1}=\mathrm{0}\:\mid\:\mathrm{x}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{1}\:\mid\:\:\mathrm{x}=\mathrm{2} \\ $$$$\mathrm{Solution}\:\mathrm{Set}:\left\{\mathrm{1},\mathrm{2}\right\}=\left\{\mathrm{n}\mid\mathrm{n}\in\mathbb{N}\wedge\mathrm{n}<\mathrm{3}\right\} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{x}−\mathrm{1}=\mathrm{0}\:\mid\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{1}\:\mid\:\mathrm{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\mathrm{x}=\mathrm{1},\:\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\pm\mathrm{i}\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{Solution}\:\mathrm{Set}=\mathrm{B} \\ $$
Commented by tawa tawa last updated on 15/May/17
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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