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Determine the value(s) of K that makes 9w^2-kw+36 factorable.Please explain how you...
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`P = 9w^2-kw+36`
For P to be factorable `Delta >= 0 `
For a function of `ax^2+bx+c` Delta is defined as;
`Delta = b^2-4ac`
So for this question;
`Delta = (-k)^2-4*9*36`
`k^2 - 36^2 >=0`
When `k<-36` suppoose k = -40 then `(k+36)(k-36) >0`
When `k = -36` then (k+36)(k-36) = 0
When` -36<k<36` suppoose k = 0 then `(k+36)(k-36) <0`
When `k = 36` then `(k+36)(k-36) = 0`
When `k>36` suppoose k = 40 then `(k+36)(k-36) >0`
So `(k+36)(k-36) >=0 ` only when ;
k<=-36 and k>=36
In advanced notation we could write this as;
`k in (-oo,-36]U[36,+oo)`
So `9w^2-kw+36` factorable when k<=-36 and k>=36.
So ` ` factorable when `k in (-oo,-36]U[36,+oo)`
Posted by jeew-m on October 21, 2012 at 4:02 PM (Answer #1)
High School Teacher
Determine the values of k so that `9w^2-kw+36` is factorable:
There are multiple answers to this question depending on the set of numbers you are factoring over:
(1) Complex numbers: this will always factor. The fundamental theorem of algebra guarantees that every polynomial factors into linear factors.
(2) Real numbers: the analysis by jeew-m is correct. Any value of k where the determinant is greater than or equal to zero will allow factoring over the reals.
(3) Rationals: Let p and q be numbers so that their product is 324 . Then any k=p+q will be factorable over the rationals.
For example, let p=27 and q=12. Then k=39.
Here the roots are 3 and `4/3`
The possibilities are 325,164,111,85,60,45,39,36
(4) Integers: Here the discriminant must not only be positive, but also a perfect square number. k=45 or k=36 are the possibilities.
Posted by embizze on October 21, 2012 at 8:30 PM (Answer #2)
Valedictorian, Quiz Taker, Super Tutor, Tutor, Dean's List
This type of sum is not thought in class 7.
Posted by astrosonuthird on October 22, 2012 at 7:26 AM (Answer #4)
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