Question

I am trying to find the factors of 3r^2+3=-1. I already have the solution. I tried setting it to zero, but it did not work.

Answer 1

For this problem, we are given a quadratic equation, and we need to determine its roots.

The equation is given below:

`3r^2+3=-1`

We can simplify the equation as shown below:

`\begin{gathered} 3r^2+3+1=-1+1\\ \\ 3r^2+4=0\\ \\ r^2=-(4)/(3)\\ \\ r=\pm\sqrt{-(4)/(3)}\\ \\ r=\pm(2i)/(√(3))\\ \\ r=\pm(2i√(3))/(3) \\ \end{gathered}`

Now we need to check whether the solutions are correct, for that we need to replace the root in the place of "r", and the equation should be equal to 0. We have:

`\begin{gathered} 3\cdot((2i√(3))/(3))^2+4=0\\ \\ (3\cdot4\cdot(-1)\cdot3)/(9)+4=0\\ \\ (9\cdot(-4))/(9)+4=0\\ \\ -4+4=0\\ \\ 0=0 \end{gathered}`

Therefore, we can conclude that the solution is correct.

We can represent any polynomial as the product of its zeros, for a quadratic term, we would have the following solution:

`(x-a)(x-b)`

Where "a" and "b" are the roots of the polynomial, therefore for this case we have:

`\begin{gathered} 3r^2+4=0\\ \\ (r-(2i√(3))/(3))(r_+(2\imaginaryI√(3))/(3))=0 \\ (1)/(3)(3r-2i√(3))(3r+2i√(3))=0 \end{gathered}`