Question

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

Answer 1

Answer:
The solution set is:
{$x = \frac{1}{6} + \frac{ \sqrt{13} }{6} \: or \: x = \frac{1}{6} - \frac{ \sqrt{13} }{6}$}

Step-by-step 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

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

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

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

$\implies x = \frac{1}{6} + \frac{ \sqrt{13} }{6} \: or \: x = \frac{1}{6} - \frac{ \sqrt{13} }{6}$

Answer 2

Answer:

$x = \frac{1}{6} + \frac{ \sqrt{13} }{6} \:$ or $\: x = \frac{1}{6} - \frac{ \sqrt{13} }{6}$

Step-by-step 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:

$x = \frac{1}{6} + \frac{ \sqrt{13} }{6} \:$ or $\: x = \frac{1}{6} - \frac{ \sqrt{13} }{6}$