3x^2-2x-1=0?

Question

how do you solve for x?

Answer 1

Reverse Foil
(3x + 1) * (x - 1) = 0
Thus, when each factor is set to zero...
x -1 =0
and
3x + 1 = 0
The roots simply become +1 and - 1/3

Answer 2

3x 2 2x 1

Answer 3

There are a few ways. The most general is the quadratic formula:

x = [ -b +/- sqrt( b^2 - 4ac ) ] / [ 2a ]

For this equation, a = 3, b = -2, and c = -1
b^2 - 4ac = (-2)^2 - (4)(3)(-1) = 4 + 12 = 16
sqrt(16) = 4

x = [ 2 +/- 4 ] / [ 6 ]
x = 6/6 = 1, or -2/6 = -1/3

Check:
3(1^2) - 2(1) - 1 =? 0
3 - 2 -1 =? 0
0 =? 0, yes, so this answer checks okay

Checking x = -1/3 is left for you to do.

Answer 4

x = [ 2 ± √(4 + 12) ] /6
x = [2 ± 4] / 6
x = 6/6 , x = - 2/6
x = 1 , x = - 1/3

Using factors:
(3x + 1).(x - 1) = 0
x = - 1/3 , x = 1 (as above)

Answer 5

factor the equation

two numbers that multiply to -3 and add to -2

the two numbers are -3 and 1

because the leading coefficient is not 1, you need to replace the middle term with the two numbers we found.

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

factor by groups
(3x^2 - 3x) + (1x - 1) = 0

factor out GCF
3x(x - 1) + 1(x - 1) = 0

factor out x - 1
(x - 1) (3x + 1) = 0

x = 1 or -1/3

Answer 6

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to find the discriminant we use the formula b^2-4ac, which is the inside of the root of the quadratic equation. When replacing a with 3, b with 2 and c with 1 we get (2)^2-4(3)(1)= -8. Since the answer is negative it is an imaginary number, so the answer is A two imaginary solutions. The reason there are two imaginary is because when taking the square root you have a positive square root and a negative one. Hope this helps...

Answer 7

The usual way. (3x+1)(x-1)=0

Answer 8

discrim is the part under the sqare root when using the quadratic formula: b^2 - 4ac: here is it negative, so you have A. 2 imaginary solutions. square roots of negative numbers do this! good luck!