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

2 Answers
Nov 22, 2017

"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

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