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

Answer 1

Eva Abramson

#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) :}#

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Answer 2

Ethan Adkins

To solve the equation 12x^2 + 2x = 0, you can factor out the common factor of 2x, giving you 2x(6x + 1) = 0. Then, set each factor equal to zero and solve for x. So, 2x = 0 gives x = 0, and 6x + 1 = 0 gives x = -1/6. Therefore, the solutions are x = 0 and x = -1/6.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.