Question

determine the number of real solutions 2x^2+12x+18=0

Answer 1

By the fundemental Theorem of algebra
The number of solutions can be determined by the degree of the polynomial

here we have a 2nd degree polynomial
that means there should be 2 solutions
let's find them
2x^2+12x+18

1st of all we will have to find the factors of p = 2 and q = 18

\[p: \pm1 , \pm 2\]
\[q: \pm1 , \pm2 , \pm3 , \pm18\]

q/p would be the possible solutions
\[\frac{ q}{ p } : \pm1, \pm2, \pm3, \pm 18 , \pm1/2 , \pm3/2 , \pm9\]

possible positive solutions: 2x^2+12x+18
is zero (because the sign hasn't change even for once)

Possible negative solutions
2x^2+12x+18
we will have to make each x = -x
so after simplifying we would have \
2x^2 - 12x + 18
the sign here changed once so we would have 1 negative solution

Compolex possible solutions:

+ 0 0
- 1 0
Complex 1 2

so now let's try the possible solutions (negative once) and see which would work as a solution