How do you solve #3x^2+7x+4=0#?

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Answer 1 · 2016-03-22T00:19:37

See explanation for a few methods...

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Quick Method

Notice that if you invert the sign of the term of odd degree then the sum of the coefficients is zero. That is: #3-7+4 = 0#

We can deduce that #x=-1# is a root and #(x+1)# a factor:

#0 = 3x^2+7x+4=(x+1)(3x+4)#

So the other root is #x = -4/3#

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AC Method

Look for a pair of factors of #AC = 3*4 = 12# with sum #B=7#.

The pair #3, 4# works.

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

#0 = 3x^2+7x+4#

#=3x^2+3x+4x+4#

#=(3x^2+3x)+(4x+4)#

#=3x(x+1)+4(x+1)#

#=(3x+4)(x+1)#

Hence roots #x=-4/3# and #x = -1#

Use the difference of squares identity too:

#a^2-b^2 = (a-b)(a+b)#

with #a = (6x+7)# and #b=1#

First multiply the equation by #2^2*3 = 12# to cut down on the fractions involved:

#0 = 3x^2+7x+4#

becomes

#0 = 36x^2+84x+48#

#=(6x+7)^2-49+48#

#=(6x+7)^2-1^2#

#=((6x+7)-1)((6x+7)+1)#

#=(6x+6)(6x+8)#

#=(6(x+1))(2(3x+4))#

#=12(x+1)(3x+4)#

Hence roots #x=-1# and #x=-4/3#