Question

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

Answer 1

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

Explanation:

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`