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

3 Answers
Jan 15, 2018

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 :)

Jan 15, 2018

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

Jan 15, 2018

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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Tony BTony B