Question 610240
{{{3x^2 + 7x > -12}}}
Like quadratic equations, you want one side to be zero and then factor the other side. Adding 12 to each side we get:
{{{3x^2 + 7x + 12 > 0}}}
Then we factor. But this won't factor!<br>
We can use the Quadratic Formula to tell us something helpful:
{{{x = (-(7) +- sqrt((7)^2 - 4(3)(12)))/2(3)}}}
Simplifying:
{{{x = (-(7) +- sqrt(49 - 4(3)(12)))/2(3)}}}
{{{x = (-(7) +- sqrt(49 - 144))/2(3)}}}
{{{x = (-(7) +- sqrt(-95))/2(3)}}}
At this point, with the negative in the square root, we can stop. The formula is trying to tell us x value(s) that make {{{3x^2+7x+12}}} equal to zero. Since we cannot have a negative inside a square root, this tells us that {{{3x^2+7x+12}}} is <i>never</i> equal to zero.<br>
If {{{3x^2+7x+12}}} is never equal to zero, it must always be positive or always be negative. We can quickly see that when x = 0, {{{3x^2+7x+12}}} will be a 12. 12 is positive. Since {{{3x^2+7x+12}}} is always positive or negative and since we have found a case where it is positive, {{{3x^2+7x+12}}} must always be positive. So the solution to
{{{3x^2 + 7x + 12 > 0}}}
which says that {{{3x^2 + 7x + 12}}} is positive, is:
All real numbers. In other words, no matter what number you use for x, {{{3x^2 + 7x + 12}}} will work out to be positive.