How do you solve using the completing the square method #x^2 + 2x - 15 = 0#?

2 Answers
Apr 4, 2016

Answer:

The solutions are:
#color(green)(x =3# or # color(green)(x = -5#

Explanation:

#x^2 + 2x - 15 =0 #

#x^2 + 2x = 15 #

To write the Left Hand Side as a Perfect Square, we add 1 to both sides:

#x^2 + 2x + 1 = 15 + 1#

#x^2 + 2 * x *1 + 1^2 = 16#

Using the Identity #color(blue)((a+b)^2 = a^2 + 2ab + b^2#, we get

#(x+1)^2 = 16#

#x + 1 = sqrt16# or #x + 1 = -sqrt16#

#color(green)(x = 4 - 1 =3# or # color(green)(x = 4 - 1 = -5#

Apr 4, 2016

Answer:

#color(blue)(=> +x=-1+-4)#

#color(brown)(x= +3 " or "-5)#

Explanation:

The equation in standard form is:
#" "color(green)(y=ax^2+bx+c)#

The equation in vertex form is:
#" "color(blue)(y=a(x+b/(2a))^2+c-[(b/2)^2])#

#color(brown)("The "-[(b/2)^2]" corrects the error produced by" (b/(2a))^2#

They are different versions of the same thing!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Solving your question")#

#a = 1" ; "b=2" ; "c=-15# giving

#y= (x+1)^2-15-1#

#color(blue)(y=(x+1)^2-16)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set the vertex form equation as equal to zero

#=>0=(x+1)^2-16#

#=> (x+1)^2=+16#

Taking the square root of both sides

#=> +-(x+1)=+-sqrt(16)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the case "-(x+1))#

#=> -x-1=+-4#

#=>-x=+1+-4#

Multiply both sides by (-1)

#color(blue)(=> +x=-1+-4)#

#color(brown)(x= +3 " or "-5)#

'....................................................................

#color(blue)("Consider the case "+(x+1))#

#=>x=-1+-4#

#color(brown)("Same as for "-(x+1))#
'...................................................