Question Number 20308 by Tinkutara last updated on 25/Aug/17
$$\mathrm{For}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:{k},\:\left({x}\:+\:{y}\:+\:{z}\right)^{\mathrm{2}} \:+ \\ $$$${k}\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{resolved}\:\mathrm{into} \\ $$$$\mathrm{linear}\:\mathrm{rational}\:\mathrm{factors}? \\ $$
Answered by Tinkutara last updated on 28/Aug/17
![(x+y+z)^2 +k(x^2 +y^2 +z^2 ) =x^2 +y^2 +z^2 +2xy+2yz+2zx+k(x^2 +y^2 +z^2 ) =z^2 [((x/z))^2 +((y/z))^2 +1+((2xy)/z^2 )+((2y)/z)+((2x)/z)+k((x/z))^2 +k((y/z))^2 +k] =z^2 [X^2 +Y^2 +1+2XY+2Y+2X+kX^2 +kY^2 +k] =z^2 [(k+1)X^2 +(k+1)Y^2 +2XY+2Y+2X+(k+1)] Δ=(k+1)(k+1)(k+1)+2(1)−(k+1)−(k+1)−(k+1)=0 =(k+1)^3 −3(k+1)+2=0 ⇒k=0,−3](https://www.tinkutara.com/question/Q20573.png)
$$\left({x}+{y}+{z}\right)^{\mathrm{2}} +{k}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right) \\ $$$$={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{2}{yz}+\mathrm{2}{zx}+{k}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right) \\ $$$$={z}^{\mathrm{2}} \left[\left(\frac{{x}}{{z}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{z}}\right)^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{2}{xy}}{{z}^{\mathrm{2}} }+\frac{\mathrm{2}{y}}{{z}}+\frac{\mathrm{2}{x}}{{z}}+{k}\left(\frac{{x}}{{z}}\right)^{\mathrm{2}} +{k}\left(\frac{{y}}{{z}}\right)^{\mathrm{2}} +{k}\right] \\ $$$$={z}^{\mathrm{2}} \left[{X}^{\mathrm{2}} +{Y}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{XY}+\mathrm{2}{Y}+\mathrm{2}{X}+{kX}^{\mathrm{2}} +{kY}^{\mathrm{2}} +{k}\right] \\ $$$$={z}^{\mathrm{2}} \left[\left({k}+\mathrm{1}\right){X}^{\mathrm{2}} +\left({k}+\mathrm{1}\right){Y}^{\mathrm{2}} +\mathrm{2}{XY}+\mathrm{2}{Y}+\mathrm{2}{X}+\left({k}+\mathrm{1}\right)\right] \\ $$$$\Delta=\left({k}+\mathrm{1}\right)\left({k}+\mathrm{1}\right)\left({k}+\mathrm{1}\right)+\mathrm{2}\left(\mathrm{1}\right)−\left({k}+\mathrm{1}\right)−\left({k}+\mathrm{1}\right)−\left({k}+\mathrm{1}\right)=\mathrm{0} \\ $$$$=\left({k}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}\left({k}+\mathrm{1}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{0},−\mathrm{3} \\ $$
Commented by ajfour last updated on 28/Aug/17
$${Find}\:{the}\:{coefficients}\:{if}\:{k}=−\mathrm{3}\:. \\ $$