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Now, I tried to figure this out. There aren't any denominators, so I don't have to worry about that. I can't factor this, so I used the quadratic formula, but my x values came to imaginary numbers, so does this mean there aren't any zeros, and therefore no solution? Help...

2006-09-14 17:07:51 · 4 answers · asked by *_Eyez Like Yourz_* 1 in Education & Reference Higher Education (University +)

4 answers

yes, whenever x turns out to be an i number, it implies on the graph, the function never crosses the x-axis, thus yield no real solution.

2006-09-14 17:11:05 · answer #1 · answered by Anonymous · 6 0

What you are looking for is both of the x-intercepts....Whether you try the quadratic formula or factor, it is not possible.

the y-intercept is at 8.............and because it is X squared the parabola goes upward...........it never touches the x-axis....
the answer is all real numbers satisfy that inequality........any number you pick for X.........when you add 8 will be greater than zero ........

are you sure its not negative x squared? Usually in these types of problems you are trying to find where the x-intercepts are

2006-09-15 00:27:07 · answer #2 · answered by mastahmind211 1 · 0 0

If you think about that - any x^2 would be positive... way larger than 3x (even if x is a negative) +8 --> that is ALWAYS going to be greater than zero...

So.... every x would be a solution of the inequality...

Whatever x you pick - it will always be positive - therefore greater than zero.

2006-09-15 00:19:24 · answer #3 · answered by linen 2 · 0 0

No real zeroes. Thus it is a parabola lying entirely above the x-axis, so all values of x satisfy the inequality. x = the set of real numbers.

2006-09-15 00:20:56 · answer #4 · answered by banjuja58 4 · 0 0

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