diana25
diana25
10-08-2016
Mathematics
contestada
How to solve number 1?
Respuesta :
dalendrk
dalendrk
10-08-2016
[tex]E(x)=5(x-1)(x+2)-(x+2)^2\\\\1)\ 5(x^2+2x-x-2)-(x^2+2x\cdot2+2^2)\\=5(x^2+x-2)-(x^2+4x+4)\\=5x^2+5x-10-x^2-4x-4\\=4x^2+x-14\neq4x^2+x+1\\\\2)\ E(x)=1\ for\ E(x)=4x^2+x+1\\\\4x^2+x+1=1\ \ \ \ |subtract\ 1\ from\ both\ sides\\4x^2+x=0\\x(4x+1)=0\iff x=0\ \vee\ 4x+1=0\\x=0\ \vee\ 4x=-1\\\boxed{x=0\ \vee\ x=-0.25}[/tex]
[tex]3)\ H(x)=9x^2-(2x+1)^2\\\\a.\ H(x)=(3x)^2-(2x+1)^2=[3x-(2x+1)](3x+2x+1)\\=(3x-2x-1)(5x+1)=(5x+1)(x-1)\\\\b.\ H(x)=0\\\\(5x+1)(x-1)=0\iff5x+1=0\ \vee\ x-1=0\\5x=-1\ \vee\ x=1\\\boxed{x=-0.2\ \vee\ x=1}[/tex]
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