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!

Guest Feb 16, 2021

#1**+1 **

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?

=^._.^=

catmg Feb 16, 2021

#2**+1 **

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?

Guest Feb 16, 2021

#3**+1 **

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

=^._.^=

catmg
Feb 16, 2021

#4**+1 **

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....

Melody Feb 16, 2021