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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
+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
+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
+1973
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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
+113779
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Melody  Feb 16, 2021
#7
+1

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

Guest Feb 16, 2021
#8
+113779
0

Nice sleuthing guest,

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

Melody  Feb 16, 2021
#4
+113779
+1

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