contestada

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

Respuesta :

kudzordzifrancis
Answer:
The solution set is:
{[tex]x = \frac{1}{6} + \frac{ \sqrt{13} }{6} \: or \: x = \frac{1}{6} - \frac{ \sqrt{13} }{6}[/tex]}

Step-by-step explanation:

Comparing the general equation,

[tex]a{x}^{2} + bx + c = 0[/tex]

to

[tex]3{x}^{2} - x - 1 = 0[/tex]

It can be testified that,

[tex]a = 3[/tex]

[tex]b = - 1[/tex]

and

[tex]c= - 1[/tex]

The quadratic formula is given by:

[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} [/tex]

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

[tex]x = \frac{ - ( -1) \pm \sqrt{ { (- 1)}^{2} - 4(3)( - 1) } }{2(3)}[/tex]

[tex]\implies x = \frac{1\pm \sqrt{ { 1} +12 } }{6}[/tex]

[tex]\implies x = \frac{1\pm \sqrt{ {13 } } }{6}[/tex]

[tex]\implies x = \frac{1}{6} + \frac{ \sqrt{13} }{6} \: or \: x = \frac{1}{6} - \frac{ \sqrt{13} }{6} [/tex]
alinakincsem

Answer:

[tex]x = \frac{1}{6} + \frac{ \sqrt{13} }{6} \:[/tex] or [tex]\: x = \frac{1}{6} - \frac{ \sqrt{13} }{6}[/tex]

Step-by-step explanation:

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

[tex]3x^2-x-1=0[/tex]

Since we cannot factorize it, we will use the quadratic formula [tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}[/tex]

Substituting the given values in the above formula to get:

[tex]x = \frac{ - ( -1) \pm \sqrt{ { (- 1)}^{2} - 4(3)( - 1) } }{2(3)}[/tex]

[tex]x = \frac{1\pm \sqrt{ { 1} +12 } }{6}[/tex]

[tex]x = \frac{1\pm \sqrt{ {13 } } }{6}[/tex]

So the roots are:

[tex]x = \frac{1}{6} + \frac{ \sqrt{13} }{6} \:[/tex] or [tex]\: x = \frac{1}{6} - \frac{ \sqrt{13} }{6}[/tex]

Otras preguntas