solve the rational equation below?

Question

1. z^2/8-7z/8=1

2. n^2/2-5n/6-2=0

3. x+1/2x=x-1/2x-1

4. 2/m^2+3m-4=2/3m+12-m/m-1

5. given a zero: x=2, of the polynomial: x^3-4x^2+21x-34=0, find the others.

Answer 1

5. You know one factor is (x-2). The other factor will be of the form ax^2 + bx + c.

(x-2)*(ax^2 + bx + c) = ax^3 + bx^2 + cx - 2ax^2 - 2bx - 2c
group like terms...
=ax^3 + (b-2a)x^2 + (c-2b)x - 2c

so a =1
b-2a=-4 so b = -2
c-2b=21 so c = 17
and wow -2c=-34, imagine that

Solve ax^2 + bx + c=0 ie: x^2 - 2x + 17 = 0 for the other two factors...

In the above 4 equations, not gonna do the work for you, multiply each side by the LCD to make them simple (non-fractional cooefficient) quadratic equations, the LCD's are...

1. 8
2. 6
3. 2
4. (m^2)*3*(m-1)

For #4 note that m=1 or -1 would 0 a denominator and so cannot possibly be a valid solution.

Answer 2

I've made some assumptions regarding your order of operations.

1. (z^2)/8 - (7*z)/8 = 1 The denominator is common so:
(z^2-7*z)/8 = 1 Now cross multiply:
z^2-7*z = 8 Now move the 8 to the other side:
z^2-7*z-8 = 0 Now factor it:
(z+1)*(z-8) = 0 Now set each term to 0:
z+1=0 thus, z=-1 AND z-8=0 thus, z=8

2. (n^2)/2-(5*n)/6-2 = 0 Make the denominator common:
3*(n^2)/6-(5*n)/6-2*6/6 = 0 Combine them
(3*n^2-5*n-2*6)/6 = 0 Cross multiply to get rid of the denominator:
3*n^2-5*n-12 = 0 Now factor it:
(n-3)*(3*n+4) = 0 Now solve for each:
n-3=0 thus, n=3 AND 3*n+4=0 thus n=-4/3

3. (x+1)/(2*x) = (x-1)/(2*x-1) Cross Multiply:
(x+1)*(2*x-1) = (x-1)*(2*x) Now FOIL it:
2*x^2-x-2*x-1 = 2*x^2-2*x Cancel the 2*x^2 on each side:
-x+2*x-1 = -2*x Now put it all on one side:
-x+2*x+2*x-1 = 0 Combine like terms:
3*x-1 = 0 Now solve and get x=1/3

4) 2/(m^2+3*m-4) = 2/(3*m+12) - m/(m-1) Do some factoring:
2/[(m-1)*(m+4)] = 2/[3*(m+4)] - m/(m-1)
Make a common denominator for the right side:
2/[(m-1)*(m+4)]= [2*(m-1)]/[3*(m-1)*(m+4)] - [m*3*(m+4)]/[3*(m-1)*(m+4)] Now combine the right side and cancel the denominator with the left side:
2 = (2*m-2-3*m^2-12*m)/3 Cross Multiply and combine like terms:
6 = -3*m^2-10*m-2 Move the terms to the left side:
3*m^2-10*m+8 = 0 Factor it:
(3*m+4)*(m+2) = 0 Solve for each:
3*m+4 = 0 thus m=-4/3
m+2 = 0 thus m=-2

5) This polinomial forms a graph that only crosses the y-axis at one point: x=2. I'm changing the '0' to a 'y' to make the graph. So I'm afraid I don't know what 'others' you may be talking about.

Answer 3

1)z^2-7z-8=0 z=((7+-sqrt(49+32))/2
z= 8 and z= -1
2)3n^2-5n-12=0 n=(( 5+-sqrt(25+144))/6
n=3 and n= -4/3
3)x+1/2x-x+1/2x=-1 so x=-1
4) there are a lot of parenthesis missing
5)(x-2)(x^2-2x+17)=0
the second factor has complex roots
x= ((2+-sqrt(4-68)/2 = 1+-4i