How do you Identify the solutions to the equation #2x^2+5x=3# when you solve by factoring?

1 Answer
Aug 20, 2016

Answer:

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

Explanation:

Firstly we note that it is a quadratic.( has #x^2#), so we make it equal to 0.

#2x^2 +5x -3=0 " factorise"#

Find factors of 2 and 3 which subtract to give 5.
The signs will be different, there must be more positives.

As both 2 and 3 are prime numbers there are only 2 possibilities.
Cross multiply the factors:

#color(white)(xxxx)(2)" "(3)#
#color(white)(xx.x) (1" " 3)rArr" " 2xx3 = 6#
#color(white)(xx.x) (2" " 1)rArr" " 1xx1 = 1 " "6-1 =5#

we have the correct factors, now put in the correct signs.

#color(white)(xx.x)(2)" "(3)#
#color(white)(xx.x) (1" " +3)rArr" " 2xx+3 = +6#
#color(white)(xx.x) (2" "- 1)rArr" " 1xx-1 = -1 " "+6-1 =+5#

The factors are:

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

One of the factors must be 0 to give a product of 0.

#x+3 = 0 rArr x =-3#
#2x-1 = 0 rArr x = 1/2#