identify the solutions to the equation 2x^(2)+5x=3 when u solve by factoring

Question

a. 1/2, -3

b. -1/2, 3

c. 2/3, -3

d. -2/3, 3

Answer 1

First bring the 3 over to the left side of the equation to form a trinomial equal to 0.

2x² + 5x - 3 = 0

Now we have to factor the trinomial. Since the leading coefficient, 2, is prime, we know one of the binomial factors will contain a 2x and the other will contain an x, like this: (2x ± ?)(x ± ?). (If the leading coefficient is not prime, say for example 4, then it could be one of various possibilities: 2x and 2x, or x and 4x, so you would need to play around with it.)

The c term, -3, is negative, so we know one of the factors will contain a plus sign and the other a minus sign. The only factors that multiply to 3 are 1 and 3, so now we have to figure out in which binomial to put the plus or minus and the 1 or the 3. We do this by adding the outers and inners to make +5x.

(2x - 1)(x + 3) = 0          This factorization enables us to get +5x when we add the outers and inners.

Now we set each factor to 0 and solve for x.

2x - 1 = 0          x + 3 = 0

2x = 1               x = -3

x = 1/2

So the solutions to this equation are 1/2 and -3, choice A

Answer 2

Well there are a bunch of possibilities here so we have to first rearrange the equation into standard form and then look at the possible combinations for factoring:

std form: 2x^2+5x-3=0

The possible factoring solutions are (why?):

a. (2x+3)(x-1)=0 (middle term is +1)

b. (2x-3)(x+1)=0 (middle term is -1)

c. (2x+1)(x-3)=0 (middle term is -5)

d. (2x-1)(x+3)=0 (middle term is +5)

Since the middle term must +5, we are forced to use solution d. as the answer. Solving each factor for zero gives us x=1/2 and x=-3.