How do you solve #x^2-12x+5=0# by completing the square?
3 Answers
just add and subtract
Explanation:
so we have
therefore
so
hope u find it helpful :)
Explanation:
#"using the method of "color(blue)"completing the square"#
#• " the coefficient of the "x^2" term must be 1 which it is"#
#• " add/subtract "(1/2"coefficient of x-term")^2" to"#
#x^2-12x#
#rArrx^2+2(-6)xcolor(red)(+36)color(red)(-36)+5=0#
#rArr(x-6)^2-31=0#
#rArr(x-6)^2=31#
#color(blue)"take the square root of both sides"#
#rArrx-6=+-sqrt31larrcolor(blue)"note plus or minus"#
#rArrx=6+-sqrt31#
See explanation
Explanation:
For a shortcut method see https://socratic.org/s/aMzZC8RW
This is actually changing
Where
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This uses the 'perfect square' but modifies it so that it matches the given equation. A bit like the logic of
Perfect square
Compare:
Using Point(1) and Point(2) now we have:
We now have to change the standard form so that it has the same overall value of the given equation. We need to change
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Vertex
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For x-intercepts
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From
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