Menu Close

Question-143024




Question Number 143024 by Feruzbek last updated on 09/Jun/21
Answered by bramlexs22 last updated on 09/Jun/21
(1)⇔ 49−x^2  ≤ 24   ⇔ 25 −x^2 ≤0  ⇔ (x+5)(x−5)≥0  ⇔ x≤−5 ∪ x≥5  (2) 49−x^2  > 0  ⇒x^2 −49<0  ⇒−7<x<7  (3)x−2>0⇒x>2  Solution (1)∩(2)∩(3)  ⇒ 5≤x<7 or [ 5, 7)
$$\left(\mathrm{1}\right)\Leftrightarrow\:\mathrm{49}−{x}^{\mathrm{2}} \:\leqslant\:\mathrm{24}\: \\ $$$$\Leftrightarrow\:\mathrm{25}\:−{x}^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\Leftrightarrow\:\left({x}+\mathrm{5}\right)\left({x}−\mathrm{5}\right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\:{x}\leqslant−\mathrm{5}\:\cup\:{x}\geqslant\mathrm{5} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{49}−{x}^{\mathrm{2}} \:>\:\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{49}<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{7}<{x}<\mathrm{7} \\ $$$$\left(\mathrm{3}\right){x}−\mathrm{2}>\mathrm{0}\Rightarrow{x}>\mathrm{2} \\ $$$${Solution}\:\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\cap\left(\mathrm{3}\right) \\ $$$$\Rightarrow\:\mathrm{5}\leqslant{x}<\mathrm{7}\:{or}\:\left[\:\mathrm{5},\:\mathrm{7}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *