Find the roots of the function in simplest radical form: f(x) = 3x^2+6x-2

Question

f(x)=3x²+6x-2

Answer 1

We need to use the quadratic equation here, -b ± √(b²-4ac)/2a

b = 6, a = 3, and c = -2

(-6 ± √[(6)² - (4)(3)(-2)])/(2)(3) = (-6 ± √(36 + 24))/6 = -6/6 ± √60/6 = -1 ± 2√15/6 = -1 ± √15/3

So the roots in simplest radical form are x = (-1 + √15/3) and x = (-1 - √15/3)

Answer 2

For this question, we will need to use the quadratic formula: (-b ± √(b²-4ac))/2a. We plug in the values from our quadratic equation: f(x) = ax²+bx+c = 3x²+6x-2.

a=3

b=6

c=-2

First, we plug our values in: (-6 ± √(6²-4(3)(-2)))/2(3)

Next, we want to simplify: (-6 ± √(6²+24))/6

Then, we can simplify further under the root: (-6 ± √60)/6

Now, we can simplify the root: (-6 ± √(4*15))/6 -> (-6 ± 2√15)/6

Now we can simplify using the denominator: (-3 ± √15)/3

Finally, we will split the fraction so we can simplify just one more time: -3/3 ± √(15)/3 -> -1 ± √(15)/3

This gives us our roots in the simplest radical form:

-1 + √15/3

-1 - √15/3