Need help finding intergers...

[heatherwertz3]

I am looking for 2 intergers with a product of 9 and a sum of -9. I am working on a math problem that looks like this
h^2-9hs+9s^2

(h ? )(h ?)

Any help on this would be a blessing.

 

[galactus]

\(\displaystyle h^{2}-9hs+9s^{2}\)

h is the variable and s is a constant. 9s^2 is a constant and -9s is the coefficient of the h term.

To factor, you need two numbers when multiplied equal 9s^2 and when added equal -9s.

This does not factor too well. i.e. it does not have integer roots.

If we solve via the quadratic formula, we get \(\displaystyle h=\frac{3s(\sqrt{5}+3)}{2}, \;\ h=\frac{-3s(\sqrt{5}-3)}{2}\)

 

[heatherwertz3]

That 's what I was thinking. This is supposed to be Intermediate Algebra so I do not think it is meant to be difficult, but should be a simple solution. I am beginning to think that it may be a typo in my book. Thanks for the explanation.

 

[Denis]

I am looking for 2 intergers with a product of 9 and a sum of -9. I am working on a math problem that looks like this
h^2-9hs+9s^2

Your comment "that looks like this" is sort of odd: do you mean you're guessing...could the 1st 9 be a 6?

To start, remove the s's: h^2 - 9h + 9 ; you can put the s's back in later.
2 integers with +-9 as product and +-9 as sum DO NOT exist.

What you're (apparently) told is to FACTOR that expression.
IF it was h^2 - 6h + 9, then EASY, right?

Or h^2 - 9h + 20 : easy, right?

 

[heatherwertz3]

The problem is a sort of fill in the blank. It is intended to help us with factoring. It lists the equation as h^2-9hs+9s^2. It then states to find the 2 intergers to complete the equation (h ?)(h ?). And your examples are right on...very easy. Listing the answer as a prime polynomial was not an option. I was supposed to fill in the blanks. I am sorry if the way I phrased the question was confusing, but thanks for taking the time to answer .

 

[stapel]

The problem is a sort of fill in the blank... It then states to find the 2 intergers to complete the equation (h ?)(h ?)... Listing the answer as a prime polynomial was not an option.

If the assignment is rife with errors such as misspelling "integers" as "intergers", typo-ing numbers, and not allowing for valid answers to be entered, then it is probably no wonder that this exercise was difficult to describe or complete! :wink:

 

[Denis]

I am looking for 2 intergers with a product of 9 and a sum of -9.

I'm sure you realised on your own that ONLY 9*1 and 3*3 were possible; sums of 10 and 6.
So WHY did you post the problem to start with?

 

[heatherwertz3]

Because I was looking for confirmation. You don't expect to see errors in your text book!

 

[stapel]

You don't expect to see errors in your text book!

While it is unusual to see typoes at this level of study (as such texts and resources are customarily more-closely edited), it does indeed, as you have seen here, happen from time to time.

In future, kindly please provide all the necessary information "up front" (such as "this is a fill-in-the-blank exercise", "I got this answer", and "the only sensible answer isn't a listed answer option"), so as to minimize confusion, etc.

Thank you for your consideration!