How do you solve using the completing the square method x^2-5x+3=0x25x+3=0?

1 Answer
Nov 4, 2017

x=1/2(5+-sqrt13)x=12(5±13)
[x approx 4.30278 or approx 0.69722x4.30278or0.69722]

Explanation:

x^2-5x+3=0x25x+3=0

x^2-5x =-3x25x=3

Our objective here is to make the LHS into a perfect square.

Add (5/2)^2(52)2 to both sides:

x^2-5x +(5/2)^2=-3+(5/2)^2x25x+(52)2=3+(52)2

Notice that the LHS = (x-5/2)^2=(x52)2

Hence, (x-5/2)^2 = -3 +25/4(x52)2=3+254

(x-5/2)^2 = (-12 +25)/4 =13/4(x52)2=12+254=134

:. x-5/2 =+-sqrt(13/4)

x= 5/2+-sqrt13/sqrt4

= 1/2(5+-sqrt13)

x approx 4.30278 or approx 0.69722