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Question Number 98135 by Tony Lin last updated on 11/Jun/20
how to split x^2 +xy+y^2  into  ((√3)/2)(x+y)^2 +(1/2)(x−y)^2  ?
$${how}\:{to}\:{split}\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \:{into} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+{y}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)^{\mathrm{2}} \:? \\ $$
Commented by MJS last updated on 11/Jun/20
x^2 +xy+y^2 ≠((√3)/2)(x+y)^2 +(1/2)(x−y)^2   there′s something wrong with this question
$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \neq\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+{y}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{something}\:\mathrm{wrong}\:\mathrm{with}\:\mathrm{this}\:\mathrm{question} \\ $$
Commented by MJS last updated on 11/Jun/20
x^2 +xy+y^2 =p(x+y)^2 +q(x−y)^2   ⇔  (1−p−q)(x^2 +y^2 )+(1−2p+2q)xy=0  ⇒  1−p−q=0∧1−2p+2q=0  ⇒  p=(3/4)∧q=(1/4)  ⇒  (((√3)/2)(x+y))^2 +((1/2)(x−y))^2 =x^2 +xy+y^2
$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} ={p}\left({x}+{y}\right)^{\mathrm{2}} +{q}\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$$\left(\mathrm{1}−{p}−{q}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\left(\mathrm{1}−\mathrm{2}{p}+\mathrm{2}{q}\right){xy}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{1}−{p}−{q}=\mathrm{0}\wedge\mathrm{1}−\mathrm{2}{p}+\mathrm{2}{q}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\mathrm{3}}{\mathrm{4}}\wedge{q}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$$\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+{y}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \\ $$

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