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x-2-3x-4-0-x-2-16-x-2-




Question Number 147894 by mathdanisur last updated on 24/Jul/21
x^2  - 3x + 4 = 0  ⇒  x^2  + ((16)/x^2 ) = ?
$${x}^{\mathrm{2}} \:-\:\mathrm{3}{x}\:+\:\mathrm{4}\:=\:\mathrm{0}\:\:\Rightarrow\:\:{x}^{\mathrm{2}} \:+\:\frac{\mathrm{16}}{{x}^{\mathrm{2}} }\:=\:? \\ $$
Answered by mindispower last updated on 24/Jul/21
⇔x+(4/x)=3  ⇒x^2 +((16)/x^2 )+8=9
$$\Leftrightarrow{x}+\frac{\mathrm{4}}{{x}}=\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{16}}{{x}^{\mathrm{2}} }+\mathrm{8}=\mathrm{9} \\ $$
Commented by mathdanisur last updated on 24/Jul/21
Thankyou Sir
$${Thankyou}\:{Sir} \\ $$
Answered by puissant last updated on 24/Jul/21
x^2 −3x+4=0 ⇒ x^2 +4=3x  ⇒x+(4/x)=3  ⇒(x+(4/x))^2 =9  ⇒x^2 +2x(4/x)+((16)/x^2 )=9  ⇒x^2 +((16)/x^2 )=9−8=1....
$$\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{4}=\mathrm{0}\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{4}=\mathrm{3x} \\ $$$$\Rightarrow\mathrm{x}+\frac{\mathrm{4}}{\mathrm{x}}=\mathrm{3} \\ $$$$\Rightarrow\left(\mathrm{x}+\frac{\mathrm{4}}{\mathrm{x}}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\frac{\mathrm{4}}{\mathrm{x}}+\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{9} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{9}−\mathrm{8}=\mathrm{1}…. \\ $$
Commented by mathdanisur last updated on 24/Jul/21
Thankyou Sir
$${Thankyou}\:{Sir} \\ $$

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