How do you solve #2x - 6 = -5x^2# by completing the square?

1 Answer
Jul 18, 2016

Answer:

#x=-1.313# to 3 decimal places

#x=+0.914# to 3 decimal places

Explanation:

Standard form #-> y=ax^2+bx+c#

Converting the given equation to standard form gives

#y=0=5x^2+2x-6#

But #y=0# is a specific case

So for the general case we have:

#y=5x^2+2x-6#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

the process of completing the square introduces an error. It ads a term to the original equation. Thus this term must be removed. This is achieved by introducing a correction.

Let #k# be the correction value.

#color(blue)("Step 1")#

write as:#" "y=5(x^2+2/5x)-6#

Include the correction

#y=5(x^2+2/5x)-6+k#

At this stage #k=0# as we have not changed the overall values.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

Move the power from #x^2# to outside the brackets

#y=5(x+2/5x)^2-6+k larr" now "k" starts to have a value"#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

Remove the #x# from #2/5x#

#y=5(x+2/5)^2-6+k #
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#

Halve the #2/5#

#y=5(x+2/10)^2-6+k #
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is the point where we find the value of #k#
The error comes from: #5(2/10)^2#. This is an added term and needs to be removed.

So #5(2/10)^2+k=0#

#=>color(red)(k=-(5xx4)/100" "=" "-20/100" " =" " -1/5)#

So we now have:

#y=5(x+2/10)^2-6color(red)(-1/5) #

#color(white)(2/2)#

#" "color(blue)(ul(bar(|color(white)(2/2)y=5(x+2/10)^2-31/5color(white)(2/2)|)))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Set #y=0# then we have

#+31/25=(x+2/10)^2#

#=>x+2/10=+-sqrt(31/25)#

#=>x=-2/10+-sqrt(31/25)#

#=>x=-2/10+-sqrt(31)/5#

#x=-1.313# to 3 decimal places

#x=+0.914# to 3 decimal places

Tony B