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Question-30813




Question Number 30813 by mondodotto@gmail.com last updated on 26/Feb/18
Answered by ajfour last updated on 26/Feb/18
y^2 =zx   ,   5y=2(x+z) =2(70−y)  ⇒   y=20  x and z are roots of       p^2 −50p−400=0  or    (p−25)^2 =15^2   ⇒  x, z = 25±15  .
$${y}^{\mathrm{2}} ={zx}\:\:\:,\:\:\:\mathrm{5}{y}=\mathrm{2}\left({x}+{z}\right)\:=\mathrm{2}\left(\mathrm{70}−{y}\right) \\ $$$$\Rightarrow\:\:\:{y}=\mathrm{20} \\ $$$${x}\:{and}\:{z}\:{are}\:{roots}\:{of} \\ $$$$\:\:\:\:\:{p}^{\mathrm{2}} −\mathrm{50}{p}−\mathrm{400}=\mathrm{0} \\ $$$${or}\:\:\:\:\left({p}−\mathrm{25}\right)^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{x},\:{z}\:=\:\mathrm{25}\pm\mathrm{15}\:\:. \\ $$

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