SOLUTION: I need to know if I am right on this question, Find the legal values of b in the fraction. 2b^2+3b-10 ----------- b^2-2b-8 I think the answer is b=-5, -2, 2 and 4

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: I need to know if I am right on this question, Find the legal values of b in the fraction. 2b^2+3b-10 ----------- b^2-2b-8 I think the answer is b=-5, -2, 2 and 4      Log On


   



Question 6182: I need to know if I am right on this question, Find the legal values of b in the fraction.
2b^2+3b-10
-----------
b^2-2b-8
I think the answer is b=-5, -2, 2 and 4

Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
When we've got variables in the denominator, things get risky. There are certain troublemaker values for variables in the denominator that make the entire denominator equal to zero - that, we can't have in math.

So let's take just the denominator: +b%5E2+-+2b+-+8+. We know that it's factorable to +%28b+-+4%29%28b+%2B+2%29+. Now, we DON'T want b to be values that will make that denominator zero. The thing is, we must first find the troublemaker values for b. So, what values of b will make the denominator zero?

+%28b+-+4%29%28b+%2B+2%29+=+0+ <---- Just by looking at this equation, b = 4 or b = -2 are the two values that will make the denominator zero. In other words, the fraction fails if b = 4 or -2. SO, the legal values for b is the set of all real numbers EXCEPT 4 and -2. AKA, anything, as long as it's not 4 or -2.