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x-y-8-xy-7-x-2-y-2-




Question Number 9953 by konen last updated on 18/Jan/17
x−y =8  xy=7  ⇒ x^2 −y^2  =?
$$\mathrm{x}−\mathrm{y}\:=\mathrm{8} \\ $$$$\mathrm{xy}=\mathrm{7}\:\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=? \\ $$
Answered by mrW1 last updated on 19/Jan/17
(x−y)^2 =8^2   x^2 −2xy+y^2 =64  x^2 +2xy+y^2 −4xy=64  (x+y)^2 =64+4xy=64+4×7=92  x+y=±(√(92))=±2(√(23))  x^2 −y^2 =(x+y)(x−y)  =±2(√(23))×8=±16(√(23))
$$\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xy}+{y}^{\mathrm{2}} =\mathrm{64} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} −\mathrm{4}{xy}=\mathrm{64} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{64}+\mathrm{4}{xy}=\mathrm{64}+\mathrm{4}×\mathrm{7}=\mathrm{92} \\ $$$${x}+{y}=\pm\sqrt{\mathrm{92}}=\pm\mathrm{2}\sqrt{\mathrm{23}} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\left({x}+{y}\right)\left({x}−{y}\right) \\ $$$$=\pm\mathrm{2}\sqrt{\mathrm{23}}×\mathrm{8}=\pm\mathrm{16}\sqrt{\mathrm{23}} \\ $$

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