How do you solve #6x² + x = 2#?

2 Answers
May 1, 2016

Answer:

The solutions are:

#color(blue)(x= 1/2#

#color(blue)(x= -2/3#

Explanation:

#6x^2 + x = 2#

#6x^2 + x - 2 =0 #

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=6, b=1, c= - 2#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (1)^2-(4* 6 * (-2))#

# = 1 + 48 = 49#

The solutions are found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-1)+-sqrt(49))/(2*6) = (-1+-7)/12#

#x= (-1+7)/12 = 6/12 = 1/2#

#x= (-1-7)/12 = -8/12 = -2/3#

May 1, 2016

Answer:

The method of finding the solution will depend on the grade in which it is being dealt with. An alternative method is to factorise.
#x= 1/2# or #x = -2/3#

Explanation:

To solve a quadratic #(x^2)# equation, always make it equal to 0.

In simple algebra there are then three methods available to find the solution: factorise, complete the square or the quadratic formula.

#6x^2 + x - 2 = 0# factorises into #(2x-1)(3x+2) = 0#

The fact that a product of two factors results in 0 means that at least one of the factors must be equal to 0.

If #2x-1 = 0,# then #2x = 1 and x = 1/2#

If #3x+2 = 0,# then #3x = -2 and x = -2/3#