Question

Find the solution set by factoring and using the zero product rule or by using the quadratic formula

3x^2-x-1=0

enter your answer in fractional form using exact roots

Answer 1

`x = (1)/(6) + ( √(13) )/(6) \:` or
`\: x = (1)/(6) - ( √(13) )/(6)`

Explanation:

We are given the following quadratic equation and we are to find its roots:

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

Since we cannot factorize it, we will use the quadratic formula
`x = \frac{ - b \pm \sqrt{ {b}^(2) - 4ac } }{2a}`

Substituting the given values in the above formula to get:

`x = \frac{ - ( -1) \pm \sqrt{ { (- 1)}^(2) - 4(3)( - 1) } }{2(3)}`

`x = \frac{1\pm \sqrt{ { 1} +12 } }{6}`

`x = \frac{1\pm \sqrt{ {13 } } }{6}`

So the roots are:

`x = (1)/(6) + ( √(13) )/(6) \:` or
`\: x = (1)/(6) - ( √(13) )/(6)`

Answer 2

Answer:
The solution set is:
{
`x = (1)/(6) + ( √(13) )/(6) \: or \: x = (1)/(6) - ( √(13) )/(6)`}

Explanation:

Comparing the general equation,


`a{x}^(2) + bx + c = 0`

to


`3{x}^(2) - x - 1 = 0`

It can be testified that,


`a = 3`


`b = - 1`

and


`c= - 1`

The quadratic formula is given by:


`x = \frac{ - b \pm \sqrt{ {b}^(2) - 4ac } }{2a}`

We now substitute the values of a, b and c into the formula


`x = \frac{ - ( -1) \pm \sqrt{ { (- 1)}^(2) - 4(3)( - 1) } }{2(3)}`


`\implies x = \frac{1\pm \sqrt{ { 1} +12 } }{6}`


`\implies x = \frac{1\pm \sqrt{ {13 } } }{6}`


`\implies x = (1)/(6) + ( √(13) )/(6) \: or \: x = (1)/(6) - ( √(13) )/(6)`