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Solve-0-x-2-5x-7-lt-1-




Question Number 15656 by Tinkutara last updated on 12/Jun/17
Solve : 0 ≤ x^2  − 5x + 7 < 1
$$\mathrm{Solve}\::\:\mathrm{0}\:\leqslant\:{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{7}\:<\:\mathrm{1} \\ $$
Answered by ajfour last updated on 12/Jun/17
     0≤ (x−(5/2))^2 +7−((25)/4) < 1       0≤ (x−(5/2))^2 +(3/4) < 1   ⇒       ∣x−(5/2)∣ < (1/2)         (5/2)−(1/2) < x < (5/2)+(1/2)                      2 < x < 3  .
$$\:\:\:\:\:\mathrm{0}\leqslant\:\left({x}−\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{7}−\frac{\mathrm{25}}{\mathrm{4}}\:<\:\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{0}\leqslant\:\left({x}−\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:<\:\mathrm{1} \\ $$$$\:\Rightarrow\:\:\:\:\:\:\:\mid{x}−\frac{\mathrm{5}}{\mathrm{2}}\mid\:<\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:<\:{x}\:<\:\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:<\:{x}\:<\:\mathrm{3}\:\:. \\ $$
Commented by Tinkutara last updated on 12/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by arnabpapu550@gmail.com last updated on 12/Jun/17
Please try 15661 No. question
$$\mathrm{Please}\:\mathrm{try}\:\mathrm{15661}\:\mathrm{No}.\:\mathrm{question} \\ $$

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