Given /devide both sides by
then you know that when x = 2; y = 0 so just plug those in.. and there you have it...
Hello, I have a few questions that are making my head hurt.
Question 1:
if the parabola goes through the point (2,0), what is the value of a?
Question 2:
Given
a:How many possible positive roots are there?
b:How many possible negative roots are there?
c:What are the possible negative roots?
d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)
Thanks
Well...I'm pretty sure I know the answers for a,b, and c: (a:1)(b:1)(c:-\frac{1}{1} , \frac{1}{1}.
For "d" I've used synthetic substitution and I think that the answer is -1, that last one is the one that makes my head hurt.
If I'm not much mistaken, the depressed equation is: [tex]x^2+3x+1=0[\math]
when I use the quadratic equation on it, I get a gibberish answer that is rational:-2.168 approximately.
ok, I think I really went wrong somewhere!
When I do synthetic division on the equation I find that -1 gives a remainder of 0....
I remember that if there is no remainder left after doing synthetic division then the divisor is a factor of the equation.
Now I am confused.....where did I mess up?