1. How do I solve the following equation?   2.

3. Originally Posted by ScubaDiver
How do I solve the following equation? Let u = x^2, than the equation would be equal to the following:       Now, let's check our solutions:     4. Try finding a substitution to fit it into the form needed for the quadratic equation.

Edit: Ellatha beat me to it, but the final answer could be plus or minus either of those.  5. Oh, ok. Thankyou

Yes, the answers would be plus or minus.  6. Originally Posted by ScubaDiver
Oh, ok. Thankyou

Yes, the answers would be plus or minus.
Yes; good pickup. The reason there should be a plus or minus is because the equation you provided is one of degree four, therefore there should be four roots, and since the factoring gave us two, than the other roots should be equidistant from those roots with respect to the axis of symmetry being the orgin.  7. Originally Posted by Ellatha Originally Posted by ScubaDiver
How do I solve the following equation? Let u = x^2, than the equation would be equal to the following:       Now, let's check our solutions:   You forgot a couple   8. How can I make sense of the imaginary solutions.
For example, the solutions... and can be seen through a graph. They are where the parabola crosses the x axis if the equation were graphed
But what about the imaginary solutions? I know what a complex plane is. Can I graph the function on a complex plane somehow and show the imaginary solutions are where the function crosses a certain point or line.  9. Look at some of the graphs of such functions on Wikipedia to see one possibility. Generally speaking though, there are 4 dimensions to worry about, so there's no straightforward way of doing it.  10. Actually, there are only three axis to worry about, so the equation can be graphed using 3 dimensions. x and y representing the domain, and z the range.  11. Originally Posted by ScubaDiver
How can I make sense of the imaginary solutions.
For example, the solutions... and can be seen through a graph. They are where the parabola crosses the x axis if the equation were graphed
But what about the imaginary solutions? I know what a complex plane is. Can I graph the function on a complex plane somehow and show the imaginary solutions are where the function crosses a certain point or line.
Sure, just plot the points as ordered pairs a + ib is the point (a.b) and A =ib is the point (a,-b).

Complex numbers are really just ordered pairs of real numbers with arithmetic given by

(a,b) + (c.d) = (a+b,c+d)

(a.b)(c,d) = (ac - bd, ad +bc)  Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement