How do you solve #12x^2 + 2x = 0#?

Question

4085 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-01T23:52:18

#x_1 = 0#, #x_2 = -1/6#

You can solve this quadratic by factoring it to the form

#2x(6x + 1) = 0#

The product of two distinct terms is equal to zero if either one of those terms is equal to zero, so you have

#2x = 0# or #(6x+1) = 0#

The solutions to these equations are

#2x = 0 => x = color(green)(0)#

and

#6x+1 = 0 => x = color(green)(-1/6)#

Alternatively, you could use the general quadratic form

#color(blue)(ax^2 + bx + c = 0)#

and recognize that #c=0#, which implies that the quadratic formula

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)#

is reduced to

#x_(1,2) = (-b +- sqrt(b^2 + 4 * a * 0))/(2a) = (-b +- b)/(2a)#

In your case, #a=12# and #b=2#, so the two solutions will once again be

#x_(1,2) = (-2 +- 2)/(24) = {(x_1 = (-2 +2)/24 = 0), (x_2 = (-2 -2)/24 = -1/6) :}#