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dangina [55]
3 years ago
9

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

Mathematics
2 answers:
-Dominant- [34]3 years ago
5 0
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}
pentagon [3]3 years ago
4 0

<u>Answer:</u>

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

<u>Step-by-step explanation:</u>

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}

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