How do you solve #6x^2+7x+2=0#?

2 Answers
Dec 11, 2016

Answer:

#x = -1/2# and #x = -2/3#

Explanation:

You need to factor this quadratic by playing with multipliers for 6 (1x6, 2x3, 3x2, 6x1) and multipliers for 2 (1x2, 2x1):

#(2x + 1)(3x + 2) = 0#

Now we can solve each term for #0#:

#2x + 1 = 0#

#2x + 1 - 1 = 0 - 1#

#2x = -1#

#(2x)/2 = -1/2#

#x = -1/2#

and

#3x + 2 = 0#

#3x + 2 - 2 = 0 - 2#

#3x = -2#

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

#x = -2/3#

Dec 11, 2016

Answer:

#x = -1/2" "# or #" "x = -2/3#

Explanation:

Given:

#6x^2+7x+2 = 0#

Use an AC method:

Find a pair of factors of #AC = 6*2 = 12# with sum #B = 7#.

The pair #4, 3# works.

Use this pair to split the middle term and factor by grouping:

#0 = 6x^2+7x+2#

#color(white)(0) = 6x^2+4x+3x+2#

#color(white)(0) = (6x^2+4x)+(3x+2)#

#color(white)(0) = 2x(3x+2)+1(3x+2)#

#color(white)(0) = (2x+1)(3x+2)#

Hence zeros: #x = -1/2# and #x = -2/3#