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

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Answer 1 · 2016-08-01T18:46:48

$x = \frac{1}{3}, or x = - \frac{5}{2}$ $x = 1/3 , orx = -5/2$

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

$6 x^{2} + 13 x = 5$ $color(red)(6)x^2 + 13x = 5$

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. $a$ $color(red)a$ 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.

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

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

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

How do you solve 6x^2 + 13x = 56x2+13x=56x^2 + 13x = 5 by completing the square?