Question

How do you solve `x^2-12x+5=0` by completing the square?

Answer 1

just add and subtract

Explanation:

so we have
`x^2 - 12x + 5 = 0`
`x^2 - 12x + 36 - 36 + 5 = 0`
`(x+6)^2 -31 = 0`
therefore
`x + 6 = +-31^(1/2)`
so
`x = +-31^(1/2) - 6`
hope u find it helpful :)

Answer 2

`x=6+-sqrt31`

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`

Answer 3

See explanation

Explanation:

For a shortcut method see https://socratic.org/s/aMzZC8RW
This is actually changing

`y=ax^2+bx+c` into

`y=a(x+b/(2a))^2+k+c`

Where `a(b/(2a))^2+k=0`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("The more formal approach")`

This uses the 'perfect square' but modifies it so that it matches the given equation. A bit like the logic of `"something"-x+x="something"`

Perfect square `->(a+b)^2=a^2+2ab+b^2`

Compare:

`a^2+2ab+b^2=0" "..."Standard form"`
`x^2-12x+5=0" "..."Given equation"`

`a^2=x^2=>a=x" "......"Point(1)"`

`2ab->2xb->-12x => b=-6" "....."Point(2)"`

Using Point(1) and Point(2) now we have:

`a^2 +2(a)(color(white)("dd")b)color(white)(".")+color(white)("dd")b^2color(white)("dd")=0" Standard form"`
`x^2+2(x)(-6)+(-6)^2=0" Standard form"`
`x^2color(white)("ddd")-12xcolor(white)("dd")+color(white)("dd")5color(white)("d.d")=0" "..."Given equation"`

We now have to change the standard form so that it has the same overall value of the given equation. We need to change `+(-6)^2` so that it becomes `+5. color(white)("d") ->36-31=+5`. So the modified standard form is:

`[a^2color(white)()+color(white)("ddd")2abcolor(white)("dd.")+color(white)("dd")b^2color(white)("dd")] -31=0`
`[x^2+2(x)(-6)+(-6)^2]-31=0`

`(a+b)^2-31=0`
`(xcolor(red)(-6))^2color(green)(-31)=0 larr" Vertex form (completing the square)"`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`x_("vertex")=(-1)xx(color(red)(-6)) = +6`
`y_("vertex")=color(green)(-31)`

Vertex `->(x,y)=(6,-31)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For x-intercepts

`x-6=+-sqrt(31)`

`x=6+-sqrt(31)" "` Exact values

`x=11.57 and x=0.42`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From `y=ax^2+bx+c` we have:
`y_("intercept")-> c =5`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~