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Question Number 165545 by HongKing last updated on 03/Feb/22

If 13x^2 +5y^2 +9z^2 +1=4x-6xy-12yz Find (x+y+z)(xy+xz+yz)=?

$$\mathrm{If} \\ $$$$\mathrm{13x}^{\mathrm{2}} +\mathrm{5y}^{\mathrm{2}} +\mathrm{9z}^{\mathrm{2}} +\mathrm{1}=\mathrm{4x}-\mathrm{6xy}-\mathrm{12yz} \\ $$$$\mathrm{Find} \\ $$$$\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{xy}+\mathrm{xz}+\mathrm{yz}\right)=? \\ $$

Commented by MJS_new last updated on 04/Feb/22

if x, y, z ∈C it′s not unique: we can choose two variables and solve for the third one.

$$\mathrm{if}\:{x},\:{y},\:{z}\:\in\mathbb{C}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{unique}:\:\mathrm{we}\:\mathrm{can}\:\mathrm{choose} \\ $$$$\mathrm{two}\:\mathrm{variables}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:\mathrm{the}\:\mathrm{third}\:\mathrm{one}. \\ $$

Commented by som(math1967) last updated on 04/Feb/22

yes sir

$${yes}\:{sir} \\ $$

Commented by HongKing last updated on 06/Feb/22

thank you dear Sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

Answered by som(math1967) last updated on 03/Feb/22

(4x^2 −4x+1)+(9x^2 +6xy+y^2 ) +(4y^2 +12yz+9z^2 )=0 (2x−1)^2 +(3x+y)^2 +(2y+3z)^2 =0 ⇒(2x−1)^2 =0⇒x=(1/2) (3x+y)^2 =0⇒y=−3x⇒y=−(3/2) (2y+3z)^2 =0 ⇒2y+3z=0⇒z=−((2y)/3)=1 (x+y+z)(xy+xz+yz) =((1/2)−(3/2)+1)(xy+yz+zx) =0

$$\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}\right)+\left(\mathrm{9}{x}^{\mathrm{2}} +\mathrm{6}{xy}+{y}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{12}{yz}+\mathrm{9}{z}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{3}{x}+{y}\right)^{\mathrm{2}} +\left(\mathrm{2}{y}+\mathrm{3}{z}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\left(\mathrm{3}{x}+{y}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{y}=−\mathrm{3}{x}\Rightarrow{y}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}{y}+\mathrm{3}{z}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{y}+\mathrm{3}{z}=\mathrm{0}\Rightarrow{z}=−\frac{\mathrm{2}{y}}{\mathrm{3}}=\mathrm{1} \\ $$$$\left({x}+{y}+{z}\right)\left({xy}+{xz}+{yz}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)\left({xy}+{yz}+{zx}\right) \\ $$$$=\mathrm{0} \\ $$

Commented by HongKing last updated on 06/Feb/22

thank you dear Sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

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