How do you solve #x^2-5x-10=0# by completing the square?

2 Answers
Nov 22, 2017

Answer:

"By the book":
#(x-(5+sqrt65)/2)(x-(5-sqrt65)/2)=0#

when:
#x_1=(5+sqrt65)/2~~6.53#
#x_2=(5-sqrt65)/2~~-1.53#

Explanation:

#ax^2+bx+c=0#
#x^2-5x-10=0#

#a=1#
#=>#

#(x+x_1)(x+x_2)=0#
#=>#
#(x_1+x_2)=-5=b#
#(x_1*x_2)=-10=c#

#D{-10}=(-1,10),(1,-10),(-2,5),(2,-5)#

#(-1,10):#
#-1+10=9!=-5#

#(1,-10):#
#1-10=-9!=-5#

#(-2,5):#
#-2+5=3!=-5#

#(2,-5):#
#2-5=-3!=-5#

#=>#
In that case we don't have any #ZZ# to work with, so we will move to Plan B:

#x_(1,2)={-b+-sqrt(b^2-4*a*c)}/{2*a}#
#=>#

#x_(1,2)={-(-5)+-sqrt((-5)^2+4*(1)*(-10))}/{2*(1)}=#
#={5+-sqrt(25+40)}/{2}=#
#={5+-sqrt(65)}/{2}=#
#=>#
#x_1=(5+sqrt65)/2~~6.53#
#x_2=(5-sqrt65)/2~~-1.53#

#=>#
"By the book":
#(x-(5+sqrt65)/2)(x-(5-sqrt65)/2)=0#

If we check we can easily see:

#(-(5+sqrt65)/2)*(-(5-sqrt65)/2)=(25+5sqrt65-5sqrt65-65)/4=-40/4=-10=c#

#(-(5+sqrt65)/2)+(-(5-sqrt65)/2)=(-5+sqrt65--5-sqrt65)/2=-10/2=-5=b#

Nov 22, 2017

Answer:

#x=5/2+-sqrt65/2#

Explanation:

#color(blue)"to complete the square"#

#• " ensure the coefficient of the "x^2" term is 1 which it is"#

#• " add "(1/2"coefficient of the x-term")^2" to both sides"#

#rArrx^2+2(-5/2)xcolor(red)(+25/4)-10=0color(red)(+25/4)#

#rArr(x-5/2)^2=25/4+10=65/4#

#color(blue)"take the square root of both sides"#

#rArr(sqrt((x-5/2)^2)=+-sqrt(65/4)larrcolor(blue)"note plus or minus"#

#rArrx-5/2=+-sqrt65/2#

#rArrx=5/2+-sqrt65/2#