How do you solve using the completing the square method #2x^2+8x-10=0#?

2 Answers
Feb 3, 2017

Answer:

#x=-5#
#x=1#

Explanation:

#2x^2+8x-10=2[x^2+4x-5]#

#2[x^2+4x-5]=2[(x+2)^2-2^2-5]=2[(x+2)^2-9]=#

#2(x+2)^2-18=0#

#2(x+2)^2=18#

#(x+2)^2=9#

#x+2=+-3#

#x=+-3-2#

#x=-5#

#x=1#

Feb 3, 2017

Answer:

#x=1# or #x=-5#

Explanation:

Divide both sides by #2# to get rid of the coefficient of #x^2#

#x^2+4x-5=0#

Add #5# to both sides

#x^2+4x=5#

Take half the coefficient of #x#, square it, and add to both sides

  • The coefficient of #x# is #4#
  • Half that is #2#
  • The square of #2# is #4#
  • Adding #4# to both sides gives

#x^2+4x+4=9#

Factoring the left hand side gives

#(x+2)^2=9#

The square root of both sides gives

#x+2=3# or #x+2=-3#
#x=1# or #x=-5#