Expert: Josh - 11/26/2007

Question
Find the real solution of the following equation.

0 = (x + 68)(x 2 + 99)

Answer
We need to find values of x which satisfy this equation, restricting our search to only real numbers.

Looking at the highest power of x in the first and second bracket, we have x and x square respectively. So, if we multiply out these expressions, we would obtain a cubic polynomial of degree 3. Because of this, we can expect AT MOST 3 solutions.

Easiest way to find the solution(s) is to look at each factor. The equation 0=(x+68)(x^2+99) will be satisfied, if one of the term -- either (x+68) or (x^2+99) -- equals zero.

(x+68)=0 when x=-68 ...This is the first and only real solution, because (x^2+99)=0 is equivalent to x^2=-99 which does not have real solution. i.e., we cannot square any real number (x) to get a negative number (-99).

Therefore, we have only one real solution of x for this equation, viz., x=-68.