math help?

Question

2xsquared + x - 3 = 0

my doing a math review before school starts and this problem is stumping me. the answer is supposed to be 1 according to the answer key, but it'd be great if somebody could show how it's done

Answer 1

2x^2+x-3=0
or,2x^2-2x+3x-3=0
or,2x(x-1)+3(x-1)=0
or,(x-1)(2x+3)=0
Therefore,either x-1=0 or 2x+3=0
If x-1=0,then x=1
If 2x+3=0,then x= -3/2
x=1 or -3/2 ans

Answer 2

I used the completing the square method for this, it never fails.

first step is to divide out 2:

x^2 + (1/2)x = 3/2

second step is to complete the square

x^2 + (1/4)^2x = 3/2 + 1/16

(x + 1/4)^2 = 25/16

Third take the Sqrt of both sides

x + 1/4 = +/- 5/4

fourth solve for x

x = -(1/4) + 5/4
or
x = -(1/4) - 5/4

so
x = -3/2
x = 1

If your answer can not be a negative number then x = 1 would be your answer.

Note: Completing the square can be difficult so using the quadratic equation maybe better.

Answer 3

Factor by grouping:

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

Set 2x + 3 = 0; x - 1 = 0

x=0 & x = -3/2

Answer 4

Factor it

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

let each factor = 0

x-1=0, so x=1

2x + 3 = 0, so 2x = -3, and x = -3/2

Now if your domain is only integers, or whole numbers, the fraction would be negated, and 1 would be the only answer.

Answer 5

Factor it:
(2x + 3)(x - 1) = 0, so x = 1, and x = -3/2

Now test these:
2*1 + 1 - 3 = 0 -- ok
2* (3/2)^2 - 3/2 - 3 = 2*9/4 - 3/2 - 3 = 6/2 - 3 = 0 -- ok

Answer 6

2(x^2) +x -3=0
u need to factorize "x" (middle term) into two terms such that product of the terms is 2(x^2)*(-3)=-6(x^2) and the sum of both the terms should be x...
so, we can do it like this
=> 2(x^2) -2x +3x -3 =0
take common terms outside
=> 2x(x-1) +3(x-1)=0
=> (2x+3)(x-1)=0 [ because (x-1) is common in both the brackets]

now , equating both product terms to zero separately we get,
2x+3 =0 and x-1=0
this gives
x=1 and x=(-3/2)

Answer 7

You could solve this either by factoring or by using the quadratic formula. Factoring gives:

(x -1)(2x +3) = 0; so x = 1, or x = -3/2

Using the quadratic formula is harder to depict in text, but you should get the same answer as above.

Answer 8

2x^2 + x - 3 = 0

fill in the brackets

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

either 2x + 3 = 0, if so, x = - 3/2

or x - 1 = 0, if so x = +1

Answer 9

Use the quadratic formula.

ax^2 + bx +c = 0

(-b +/- sqrt(b^2 - 4ac))/2a

In your equation,
a = 2
b = 1
c = -3

Now substitute.

x = (-1 +/- sqrt(1 - 4*2*-3))/4
x = (-1 +/- sqrt(25))/4
x = (-1 +/- 5)/4
x = 1, 6/4

Now substitute those values back into the equation to determine which works, or if both work.

2(1)^2 + 1 - 3 = 0, true
1 works.
2(6/4)^2 + (6/4) - 3 = 0, false

x = 1

Answer 10

2x^2 + x - 3 = 0

a = 2
b = 1
c = -3

x = (-b +/- sqrt(b^2 - 4ac))/(2a)
x = (-1 +/- sqrt(1 + 24))/4
x = (-1 +/- 5)/4 = -3/2 or 1