Question

Solve the quadratic equation by factoring. 2x^2+5x+3=0

Answer 1

`x_1=-(3)/(2) \\\\x_2=-1`

Explanation:

the general fromula is

`(ax+b)(cx+)d)\\\\=acx^(2) +(ad+bc)x+bd`

so we need that

`ac=2\\\\ad+bc=5\\\\bd=3`

for ac = 2

we only have 2 and 1

so let's say

`a=2\\\\c=1`

so we now have

`(2x+b)(x+d)\\\\2x^(2) +(2d+b)x+bd\\\\`

and for bd = 3

we only have 3 and 1

but we need to

2d +b = 5

so we can have

b=3

and

d=1

so the factorized form is

`(2x+3)(x+1)`

so we come back to

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

so here we have two cases

Case 1

`(2x+3)=0\\\\2x+3=0\\\\2x=-3\\\\x=-(3)/(2)`

Case 2

`(x+1)=0\\\\x+1=0\\\\x=-1`

so those are our solutions

Answer 2

-0.33 repeating

Explanation:

2x^2=4x

4x+5x+3=0

combine like terms

9x + 3 = 0

subtract 3 from both sides

9x = -3

divide 9 both sides to get the variable by itself

x = -0.33 repeating