How do you solve using the completing the square method #x^2 - 2x + 1 = 18#?

1 Answer
Feb 27, 2016

Answer:

#color(brown)("Same thing. Just a different presentation")#
#color(blue)(x_("intercept")~= + 5.24" "or" "-3.24# to 2 decimal places
#color(blue)(y_("intercept")= -17)#
#color(blue)("vertex "-> (x,y)" "->" "(1,-18))#

Explanation:

Given: #" "x^2-2x+1=18#

Subtract 18 from both sides giving:

#" "x^2-2x-17=0#.........................(1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This one is more straight forward as the coefficient of #x^2# is 1
#color(blue)("Step 1")#

Write as:

#(x^2-2x)-17=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#
#color(brown)("At this point we will change the intrinsic value of the equation so it")# #color(brown)("will need to have a correction added later.")#

Take the power to be outside the brackets

#(x-2x)^(color(red)(2))-17#

Remove the #x# from the #-2x# leaving just the #color(red)(-2)#

#(xcolor(red)(-2))^2-17#

halve the -2 that is inside the brackets so that you have #color(red)(-1)#

#(xcolor(red)(-1))^2-17#

#color(brown)("Now we add the correction")#
#color(green)("This takes the equation back to its original value so we can once again equate it to zero")#

Let #color(red)(k)# be some constant

#(x-1)^2color(red)(+k)-17=0# ................................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

#color(brown)("By expanding the brackets and comparing to the original equation")##color(brown)("we can determine the appropriate value of "k)#

Expanding equation (2)

#cancel(x^2)-cancel(2x)+1+k-cancel(17) " "=" "cancel(x^2)-cancel(2x)-cancel(17)#

#k=-1# so equation (2) becomes:

#color(blue)((x-1)^2-18=0)# .................................(3)

By comparing the values in the vertex format equation you can see how to obtain the vertex coordinates.
Multiply the constant inside the bracket by #(-1)# to get #x_("vertex")# giving 1 and #y_("vertex")# can be read directly as -18

#color(blue)("vertex "-> (x,y)" "->" "(1,-18))#

Tony B
;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#

y-intercept at #x=0#

#=>(x-1)^2-18=y" "->" "(0-1)^2-18=y#

#color(blue)(y_("intercept")= -17)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5")#

x-intercepts when y=0 so from equation (3)

#sqrt( (x-1)^2)=sqrt(18)#

#x-1=+-3sqrt(2)#

#x=1+-3sqrt(2)#

#color(blue)(x_("intercept")~= + 5.24" "or" "-3.24# to 2 decimal places