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Question Number 196360 by mathlove last updated on 23/Aug/23

Commented by mathlove last updated on 23/Aug/23

x_1 +n=?

$${x}_{\mathrm{1}} +{n}=? \\ $$

Answered by deleteduser1 last updated on 23/Aug/23

f′(x)=2x−4=0 at x_1 ⇒x_1 =2  f(x_1 )=0⇒f(2)=0⇒n=4⇒x_1 +n=6

$${f}'\left({x}\right)=\mathrm{2}{x}−\mathrm{4}=\mathrm{0}\:{at}\:{x}_{\mathrm{1}} \Rightarrow{x}_{\mathrm{1}} =\mathrm{2} \\ $$$${f}\left({x}_{\mathrm{1}} \right)=\mathrm{0}\Rightarrow{f}\left(\mathrm{2}\right)=\mathrm{0}\Rightarrow{n}=\mathrm{4}\Rightarrow{x}_{\mathrm{1}} +{n}=\mathrm{6} \\ $$

Commented by mathlove last updated on 23/Aug/23

thanks

$${thanks} \\ $$

Answered by mr W last updated on 23/Aug/23

x^2 −4x+n=(x−x_1 )^2   −4x+n=−2x_1 x+x_1 ^2   ⇒x_1 =2, n=x_1 ^2 =4  ⇒x_1 +n=2+4=6

$${x}^{\mathrm{2}} −\mathrm{4}{x}+{n}=\left({x}−{x}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$−\mathrm{4}{x}+{n}=−\mathrm{2}{x}_{\mathrm{1}} {x}+{x}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\mathrm{2},\:{n}={x}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow{x}_{\mathrm{1}} +{n}=\mathrm{2}+\mathrm{4}=\mathrm{6} \\ $$

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