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Question Number 62334 by smartsmith459@gmail.com last updated on 19/Jun/19

if α^2 +β^2 = (α+β)^2 −2αβ evaluate(α−β)

$${if}\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\:{evaluate}\left(\alpha−\beta\right) \\ $$

Answered by Kunal12588 last updated on 20/Jun/19

α^2 +β^2 =(α+β)^2 −2αβ  ⇒α^2 +β^2 −2αβ=(α+β)^2 −4αβ  ⇒α^2 +β^2 −2αβ=(α−β)^2   ⇒(α−β)=(√(α^2 +β^2 −2αβ))

$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta \\ $$$$\Rightarrow\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} −\mathrm{2}\alpha\beta=\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta \\ $$$$\Rightarrow\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} −\mathrm{2}\alpha\beta=\left(\alpha−\beta\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\alpha−\beta\right)=\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} −\mathrm{2}\alpha\beta} \\ $$

Answered by MJS last updated on 20/Jun/19

α^2 +β^2 =α^2 +2αβ+β^2 −2αβ  α^2 +β^2 =α^2 +β^2   ⇒ true for all α, β   ⇒ α−β=α−β  no other conclusion possible

$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\alpha^{\mathrm{2}} +\mathrm{2}\alpha\beta+\beta^{\mathrm{2}} −\mathrm{2}\alpha\beta \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{true}\:\mathrm{for}\:\mathrm{all}\:\alpha,\:\beta\: \\ $$$$\Rightarrow\:\alpha−\beta=\alpha−\beta \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{conclusion}\:\mathrm{possible} \\ $$

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