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

3 Answers
Jan 15, 2018

Answer:

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

Answer:

#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

Answer:

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 B