4x^2+12 = 0
I factored this out, but how do I show the final answer? Thanks!
4(x^2+3)
4(x2+3)=0
x2=-3
x=sqrt(-3)
=±(sqrt(3)*i)
where i2=-1
The roots are therefore complex.
Would iroot3 and -iroot3 be acceptable answers? Thanks!
Yes, they are acceptable answers.
When in doubt (as in an exam), the best way to check an answer is to square your proposed roots and see if you get back -3.
Thanks for the help!
To show the final answer, you need to determine the values of x that make the equation true. In this case, you have factored out common factors correctly, and now you have:
4(x^2 + 3) = 0
To find the values of x that satisfy this equation, you can set each factor to zero and solve for x individually.
Start with the first factor 4 = 0, but this is not possible since 4 is not equal to zero.
Now, set the second factor x^2 + 3 = 0 and solve for x.
x^2 + 3 = 0
To isolate x^2, subtract 3 from both sides:
x^2 = -3
To solve for x, take the square root of both sides:
x = ± √(-3)
However, there is no real number (since the square root of a negative number is not a real number) that satisfies this equation. Therefore, the equation 4x^2 + 12 = 0 has no real solutions.
In conclusion, you can show the final answer as "No real solutions" or "There are no values of x that satisfy the equation."