How do you solve #3x^2+5x-2=0#?

3 Answers
Aug 15, 2016

Answer:

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

Explanation:

Factorise to get #(3x - 1)(x + 2) = 0#

hence
#3x - 1 = 0 or x + 2 = 0#

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

Aug 15, 2016

Answer:

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

Explanation:

Find factors of 3 and 2 which subtract to make 5.

We should realise that (3x2) - 1 = 5.
Find the cross products and subtract to get 5.

#3" "1 " "rArr 1xx1 =1 #
#1" "2" "rArr 3xx2 = 6 " "(6-1 = 5)#

Add in the signs.

#3" "-1 " "rArr 1xx-1 =-1 #
#1" "+2" "rArr 3xx+2 = +6 " "(+6-1 = +5)#

#(3x-1)(x+2) = 0#
One of the factors must be 0.

If #3x-1 = 0, " then " x = 1/3#
If #x+2 = 0, " then " x = -2#

Aug 15, 2016

Answer:

There is only one way to deal with factorization. That is lots of practice.
#x=+1/3#

#x=-2#

Explanation:

Given:#" "3x^2+5x-2=0#
'............................................................................................................

Both 3 and 2 are prime numbers so they can only be the products of:

for 2#->1xx2#
for 3#->1xx3#

So without taking any notice of the signs (sorted out afterwards) we have something like: #color(red)((3x+1)(x+2) " or "(3x+2)(x+1))#
I have kept both #3x" and "x# positive.

#color(red)("Note that the signs in the above will be wrong!")#

#color(green)("Ok. Lets consider the signs by looking at the "+5x)# #color(green)("from "3x^2+5x-2)#

Notice that
#(+3x)xx(+2)=6x#
#(-1)xx x=-x#

giving #6x-x=+5x# as required

So the factor form is very likely:

#(3x-1)(x+2)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
Multiplying out the brackets
#color(blue)((3x-1))color(brown)((x+2) )" "rarr" " color(brown)(color(blue)(3x)(x+2)color(blue)(" "-1)(x+2))#

#" "=3x^2+6x" "-x-2#

#=3x^2+5x-2# as required
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So we have:

#3x^2+5x-2=0" "->" "(3x-1)(x+2)=0#

Note that anything #xx0=0#

This means that the solution is such that

#(3x-1)=0 -> x=+1/3#

#(x+2)=0->x=-2#

Tony B