1. Find all integers a for which the polynomial ax^2-8x+3=0 has an integer root.
2. The ellipse (x^2/49)+(y^2/b^2)=1. has foci at (5, 0) and (-5, 0).
2(a), find the length of the semi-major axis of this ellipse.
2(b). compute the value of B.
thank you for helping!
+2407 Hi Guest!
I have yet to learn about ellipses, so I'm sorry I can't help you on problems 2a or 2b.
However, I'm going to try question 1.
The quadratic equation is (-b+-sqrt(b^2-4ac))/2a, where the +- is the plus/minus weird symbol.
For the roots to be integers, sqrt(b^2-4ac) must be an integer.
b = -8
c = 3
sqrt(64-12a) = integer
2sqrt(16-3a) = integer
Can you try solving for them?
=^._.^=
Hello! Thank you, I understand that part! However, one question I have is that for some sqrtb^2-4ac that are integers still don't give out integer solutions because they are fractions. How would I know which ones work or not?
+2407 I'm completly stumped.
Perhaps one of the smarter people on the forum can help answer?
If you manage to figure it out, please tell me.
Sorry
=^._.^=
+118677 Please do not ask multiple questions on the same post.
You can repost Q2 on a new post, and delete if from here. (So long as no one has offered an answer on it yet)
Question 1:
1. Find all integers a for which the polynomial ax^2-8x+3=0 has an integer root.
the roots are
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {8 \pm \sqrt{64-12a} \over 2a}\\ x=\frac{4\pm \sqrt{16-3a} }{a} \)
16-3a will have to be a perfect square
a could equal 0, 4, 5
They are the only positive a values that could work
a=0 is no good - can't divide by 0
a=4 roots are fractional
a=5 one of the roots will be 1 so that works
So a=5 is the only positive value of a that will have an integer root. There might be some negative values of a that work though....
