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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!

 Feb 16, 2021
 #1
avatar+1973 
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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?

 

=^._.^=

 Feb 16, 2021
 #2
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+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?

 Feb 16, 2021
 #3
avatar+1973 
0

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
 #5
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Thank you for your help! I managed to find a couple of solutions but I'm not sure if I'm missing any lol. 

Guest Feb 16, 2021
 #6
avatar+113779 
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what answers did you find?

Melody  Feb 16, 2021
 #7
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+1

So far, I got -3, 5, and -11

Guest Feb 16, 2021
 #8
avatar+113779 
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Nice sleuthing guest,

I suspect there are infinitely many solutions if a can be negative......

Melody  Feb 16, 2021
 #4
avatar+113779 
+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....

 Feb 16, 2021

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