Some ALGEBRA 2 Questions...please help. =)?

Question

Hi, I'm having a bit of trouble on my homework....on algebra 2...The questions are:

1.
(i don't have the symbols on this keyboard so i'll write it in words)
simplify: the 3rd root of 40 mulitiplied by 4 times the 3rd root of 5. like= ^3radical40 X(times) 4^3radical5

X=multiply
2. simplify: (5^4/5 X 5^4/5) ^-10

3. simplify: the 3rd root of 2 divided by the 3rd root of 54.

4. simply: the square root of 64x^12

5. simplify: (7^3/4)^-2/9

6. write the expression in simplest form: the 3rd root (aka cubed root) of 1/5.

7. Let f(x)=4-x^2 and g(x)=2-x. find f(x) divided by g(x).

8. Let f(x)=16-x^2 and g(x)=4-x. Find (f X g)(x). (again X means multiply.)

9. Let f(x)=4-x^2 and g(x)=2-x. Find f(g(x))

10. Solve the equation and check for extraneous solutions:

the sq. root of x^2 + 5 = 3-x


OK that's the ones i'm having trouble on the most....... i KNOW it's a lot, but if you could help me with any one of them, it would be so great.

Thank you! =)

Answer 1

1) You can change the third root into exponents if that helps. You end up with 40^(1/3) * 4 * 5^(1/3). Now you simplify as you would normal exponents. 200^ (1/3) * 4

2) Start with the inside:
(25^(4/5)) ^ -10
when a power is raised to a power, all you have to do is mulitply the two together:
25^(-40/5) = 25 ^ (-8)

3) Use the same process as in problem 1:
2^(1/3) / 54^(1/3) = (1 / 27) ^ (1/3)

4) On this one you need to split up the two to get:
sqrt(64) * sqrt(x^12)
+- 8 * (x^12) ^ (1/2)
and multiply the exponents
+- 8 * x^(12/2) = +-8 x^6
(the +- 8 is plus or minus 8 since either could be squared to get 64)

5) Multiply the powers together on this one too.
7 ^ (-6 / 36) = 7 ^ (-1/6)

6) 1/5 ^ (1/3)
now to get the 5 on top, the exponent needs to be negative:
5 ^ (-1/3)

7) Start by writing it as a whole equation:
(4 - x^2) / (2 - x)
Now factor the numerator:
(2 + x)(2 - x) / (2 - x)
The (2 - x) can be cancelled to get:
2 + x

8) This one I am unfamilliar with the notation. Usually when they want you to multiply the two equations together they use f(x) * g(x). Sorry I can't help you on that.

9) What they want you to do on this one is plug the g equation into the 'x' of the f equation:
4 - (2 - x)^2
Simplify:
4 - (4 + x^2) = 4 + (-4 - x^2) = -x^2

10) On this one I dont know if the square root is only on the x^2 or on x^2 + 5.
If it is only on the x^2 then:
+-x + 5 = 3 - x
Lets start with positive x
x + 5 = 3 - x
x + 2 = -x
2 = -2x
x = -1
Now lets do negative x
-x + 5 = 3 - x
5 = 3
so that means that there is no other anwer but the one above (your extraneous solution).


If it's on both, then take the sqrt of the second side squared (they should cancel out) so you can divide one side.
But this way would get really ugly really quickly so I don't think that is what you meant.

Hope this helps :)

Answer 2

1. First multiply the numbers under like roots. You get the cube root of 200. Now break this down into a perfect cube times another number. 200 = 8 * 25, so you have the cube root of 8 times the cube root of 25. Since the cube root of 8 is 2, the answer is 2 times the cube root of 25.

2. 5^(4/5) X 5^(4/5) = 5^(8/5) because when you multiply you add exponents, so you have [ 5^(8/5) ] ^ -10
When you have a quantity to a power, you multiply exponents, so you'd multiply 8/5 and -10 and get 5^(-16). This is the same as 1/(5^16), which is a very small decimal equal to the fraction 1 / 152587890625

3. This is the cube root of 2/54 or the cube root of 1/27, which is 1/3.

4. The square root of 64 is 8. The square root of x^12 is x^6 (you divide by the index of the root). So the answer is 8x^6.

5. You multiply the exponents and get 7 ^ (-1/6), which is the same as 1 over the sixth root of 7. This could conceivably be simplified further by multiplying by the sixth root of 7 to the 5 over itself.

6. The rule they're trying to teach here is that you shouldn't have a root on the bottom of a fraction (which is why #5 could be simplified). This is the same as simplifying 1 / cube root of 5. To do this you multiply by the cube root of 25 over the cube root of 25. You get the cube root of 25 over the cube root of 125. Since the cube root of 125 is 5, the final answer is the cube root of 25 over 5.

7. This is (4 - x^2) / (2 - x). You can simplify this by factoring the top. You have [ (2 - x)(2 + x) ] / (2 - x). The factor of (2 - x) cancels, and the final answer is 1 / (2 + x).

8. This is (16 - x^2)(4 - x) which is the same as
(4 - x)(16 - x^2) Just distribute (often called FOIL). You get
64 - 4x^2 - 16x + x^3. In standard form this would be
x^3 - 4x^2 - 16x + 64

9. Wherever there is "x" in the first function, you change it to the quantity (2 - x). You get 4 - (2 - x)^2. You could just leave the answer like that, or you could simplify it to
4 - (4 - 4x + x^2) or 4x - x^2.

10. SQR(x^2 + 5) = 3 - x
Square both sides. You get x^2 + 5 = (3 - x)^2
Multiply out the right side.
You get x^2 + 5 = 9 - 6x + x^2
Cancel the x^2 terms. You have 5 = 9 - 6x
Solve. -4 = -6x ... so 2/3 = x:
Now check to make sure this answer works in the original problem.
SQR( (2/3)^2 + 5 ) = 3 - 2/3
2.3333333333 = 2.3333333333
They are equal, so this is the solution. (There is no extraneous solution.)

Answer 3

For increasing cubic binomials the time-honored formulation is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^one million + 3*a^one million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^one million + 3*x^one million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D