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

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Answer 1 · 2018-05-20T17:20:37

#x = 5+-sqrt17#

#x^2- 10x = -8#

#ax^2 +bx +c#.

To complete the square a =1 and #c = (1/2b)^2#

#c = (1/2*-10)^2 = 25#, a already is 1

we need to add the c to both sides of the equation so we don't change its value:

#x^2- 10x + c = -8 + c#

#x^2- 10x + 25 = -8 + 25#

the -5 is the #1/2b# before we squared it:

#(x -5)^2 = 17#

#sqrt((x -5)^2) = +-sqrt17#

#x -5 = +-sqrt17#

#x = 5+-sqrt17#

graph{x^2- 10x +8 [-13.16, 26.84, -18.56, 1.44]}