How do you solve using the completing the square method x^2 - 2x + 1 = 18x22x+1=18?

1 Answer
Feb 27, 2016

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

Explanation:

Given: " "x^2-2x+1=18 x22x+1=18

Subtract 18 from both sides giving:

" "x^2-2x-17=0 x22x17=0.........................(1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This one is more straight forward as the coefficient of x^2x2 is 1
color(blue)("Step 1")Step 1

Write as:

(x^2-2x)-17=0(x22x)17=0
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 2")Step 2
color(brown)("At this point we will change the intrinsic value of the equation so it")At this point we will change the intrinsic value of the equation so it color(brown)("will need to have a correction added later.")will need to have a correction added later.

Take the power to be outside the brackets

(x-2x)^(color(red)(2))-17(x2x)217

Remove the xx from the -2x2x leaving just the color(red)(-2)2

(xcolor(red)(-2))^2-17(x2)217

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

(xcolor(red)(-1))^2-17(x1)217

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

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

(x-1)^2color(red)(+k)-17=0(x1)2+k17=0 ................................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 3")Step 3

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