How do you solve #6x^2 + 13x = 5# by completing the square?

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Answer 1 · 2016-08-01T18:46:48

#x = 1/3 , orx = -5/2#

For ease of explanation - remember that the general form of a quadratic is #color(red)ax^2 + color(blue)bx + color(magenta)c#

#color(red)(6)x^2 + 13x = 5#

The constant (c) has already been moved to the RHS.

We need to make the LHS into the square of a binomial (ie a perfect square)

Step 1. #color(red)a# must be equal to 1. Divide through by 6.

#(cancel6x^2)/cancel6 + (color(blue)13x)/color(blue)6 = 5/6#

Step 2: complete the square by adding the missing third term (to both sides)

#(color(blue)b/2)^2 " "rArr ((color(blue)13)/(color(blue)6 xx2))^2 " "rArr (13/12)^2#

#x^2 + (color(blue)13x)/color(blue)6 + (13/12)^2 = 5/6 + (13/12)^2#

Step 3. Write as #(x + ....)^2#

#(x + (13)/12)^2 = 289/144#

Step 4: square root both sides.

#x + 13/12 = +-sqrt(289/144)#

#x = 17/12 -13/12 " " or x = (-17/12)- 13/12#

#x = 4/12 = 1/3" or " x = (-30)/12 =-5/2#