 #1
 655
 0
For (a) I got x = [tex]\sqrt{11}[/tex] which you cant do as you cant square root a negitive number, I cant help but feel im wrong
Attachments

12.1 KB Views: 312
Well, for that reason I would say that Bill is wrong; not you!For (a) I got x = [tex]\sqrt{11}[/tex] which you cant do as you cant square root a negitive number, I cant help but feel im wrong
What do you get when you solve this quadratic equation? The answer does not necessarily have to be a whole number, especially if it is only a short decimal. The solution to this equation is such that it can easily be written exactly.and for (b) I began be substituting y =2x  2 into the y²
x² + (2x2)(2x2) = 25
5x² 8x 21 = 0
am I doing the right thing so far. When I used the quadratic formula I got a decimal number, not a whole number and the question doesn't say anything about rounding to a degree of accuracy so I presume the answer is whole numbers