SOLUTION: I need some help with some problems, hope you could help me. I am not that good in algebra, so I need help with the following. 1. Find all numbers for which the rational expre

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I need some help with some problems, hope you could help me. I am not that good in algebra, so I need help with the following. 1. Find all numbers for which the rational expre      Log On


   



Question 327571: I need some help with some problems, hope you could help me. I am not that good in algebra, so I need help with the following.
1. Find all numbers for which the rational expression is undefined.
q^3-7q over q^2-64 (fraction).
2. Multiply & simplify by factoring. Assume that all expressions under radicals represent nonnegative numbers. 3(small three) sqrt y^4 3(small) sqrt 81y^5.
3. Use rational exponents to simplify 5sqrtx^15
4.For the following equation, state the value of the dicriminant & then describe the nature of the solutions. -16x^2+3x-3=0 What is the value of the discriminant?
5. Find the following: sqrt(a+6)^2. Assume that variables can represent any real number.

Answer by ganeshindia(11) About Me  (Show Source):
You can put this solution on YOUR website!
1) All numbers for which the rational expression
(q^3-7q)/(q^2-64)is undefined.
Solution:
The above rational expression is undefined when the denominator becomes zero. That is q%5E2+-+64=0.
That is , q%5E2+=+64.
q = + or - {sqrt of 64}
q = +8 or q= -8.
The rational expression is undefined for q = +8 and q= -8.
3) Use rational exponents to simplify 5sqrtx%5E15+.
5th+root+of+x%5E15+=+%28x%5E15%29%5E%281%2F5%29
= (x^3)^5)^(1/5)
= x%5E3
4)For the following equation, state the value of the discriminant & then describe the nature of the solutions. -16x%5E2%2B3x-3=0 What is the value of the discriminant?
Sol:
For the equation ax%5E2+%2B+bx+%2B+c+=+0, The discriminant value is b^2 - 4ac.
From the given equation, we get a = -16, b = 3 and c = -3.
Therefore,+b%5E2+-+4ac+=+3%5E2+-+4%28-16%29%28-3%29+=+9+%2B+192+=+201.
We get, b%5E2+-+4ac+%3E+0.
Therefore the roots are real, unequal and irrational.