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Question Number 161338 by akolade last updated on 16/Dec/21

Commented by cortano last updated on 16/Dec/21

  { ((16−x^2 >0⇒(x−4)(x+4)<0 ;−4<x<4)),((3x−4>0⇒x>(4/3))),((3x−4≠1 ⇒x≠(5/3))) :}  ⇔16−x^2 =(3x−4)^2   ⇔16−x^2 =9x^2 −24x+16  ⇔10x^2 −24x=0  ⇔x(10x−24)=0 ; x=((12)/5)

$$\:\begin{cases}{\mathrm{16}−{x}^{\mathrm{2}} >\mathrm{0}\Rightarrow\left({x}−\mathrm{4}\right)\left({x}+\mathrm{4}\right)<\mathrm{0}\:;−\mathrm{4}<{x}<\mathrm{4}}\\{\mathrm{3}{x}−\mathrm{4}>\mathrm{0}\Rightarrow{x}>\frac{\mathrm{4}}{\mathrm{3}}}\\{\mathrm{3}{x}−\mathrm{4}\neq\mathrm{1}\:\Rightarrow{x}\neq\frac{\mathrm{5}}{\mathrm{3}}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{16}−{x}^{\mathrm{2}} =\left(\mathrm{3}{x}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{16}−{x}^{\mathrm{2}} =\mathrm{9}{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{16} \\ $$$$\Leftrightarrow\mathrm{10}{x}^{\mathrm{2}} −\mathrm{24}{x}=\mathrm{0} \\ $$$$\Leftrightarrow{x}\left(\mathrm{10}{x}−\mathrm{24}\right)=\mathrm{0}\:;\:{x}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$

Answered by kapoorshah last updated on 16/Dec/21

(3x − 4)^2  = 16 − x^2   10x^2  − 24x = 0  2x(5x − 12) = 0  x = 0 (rejected)    x = ((12)/5)

$$\left(\mathrm{3}{x}\:−\:\mathrm{4}\right)^{\mathrm{2}} \:=\:\mathrm{16}\:−\:{x}^{\mathrm{2}} \\ $$$$\mathrm{10}{x}^{\mathrm{2}} \:−\:\mathrm{24}{x}\:=\:\mathrm{0} \\ $$$$\mathrm{2}{x}\left(\mathrm{5}{x}\:−\:\mathrm{12}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:\mathrm{0}\:\left({rejected}\right)\:\:\:\:{x}\:=\:\frac{\mathrm{12}}{\mathrm{5}} \\ $$

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