Shee2idn
Answered

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How to solve x^2-2x=3 by using completing the square.




Sagot :

[tex]x^2-2x=3 \\ \\x^2-2x-3=0 \\ \\a=1 ,\ \ b=-2, \ \ c=-3 \\ \\\Delta =b^2-4ac = (-2)^2 -4\cdot1\cdot (-3) = 4+12=16 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{2-\sqrt{16}}{2 }=\frac{ 2-4}{2}=\frac{-2}{2}=-1\\\\x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{2+\sqrt{16}}{2 }=\frac{ 2+4}{2}=\frac{6}{2}=3[/tex]
Zetidn
first you have to transpose 3
 so that it will be x^2-2X-3=0
(x-3)(x+1)
x-3=0        x+1=0
x=3           x=-1
xsub1=3     xsub2=1