Question
solve and check c-4/c-2 =c-2/c+2 - 1/2-c
Answers
GPT-4o mini
To solve the equation
\[
\frac{c-4}{c-2} = \frac{c-2}{c+2} - \frac{1}{2-c}
\]
let's first simplify the right-hand side. Notice that \(2 - c\) can be rewritten as \(-(c - 2)\). Therefore, we can rewrite the term \(\frac{1}{2 - c}\) as:
\[
-\frac{1}{c - 2}
\]
Thus, the equation becomes:
\[
\frac{c-4}{c-2} = \frac{c-2}{c+2} + \frac{1}{c - 2}
\]
Next, we need to combine the terms on the right-hand side:
\[
\text{Right side: } \frac{c-2}{c+2} + \frac{1}{c-2} = \frac{(c-2)^2 + (c+2)}{(c+2)(c-2)}
\]
The expansion of \((c-2)^2\) gives:
\[
(c-2)^2 = c^2 - 4c + 4
\]
Now add \((c + 2)\):
\[
c^2 - 4c + 4 + c + 2 = c^2 - 3c + 6
\]
So we have:
\[
\frac{c^2 - 3c + 6}{(c+2)(c-2)}
\]
Now the right-hand side becomes:
\[
\frac{c^2 - 3c + 6}{(c+2)(c-2)}
\]
Setting both sides of the equation equal:
\[
\frac{c-4}{c-2} = \frac{c^2 - 3c + 6}{(c+2)(c-2)}
\]
Now to solve this equation, we can cross-multiply:
\[
(c-4)(c+2) = (c^2 - 3c + 6)(c-2)
\]
Expanding both sides:
\[
c^2 + 2c - 4c - 8 = c^3 - 2c^2 - 3c^2 + 6c - 6
\]
This simplifies to:
\[
c^2 - 2c - 8 = c^3 - 5c^2 + 6c - 6
\]
Bringing all terms to one side gives:
\[
0 = c^3 - 5c^2 + 6c - 6 - c^2 + 2c + 8
\]
Combining like terms leads to:
\[
0 = c^3 - 6c^2 + 8c + 2
\]
Rearranging the equation:
\[
c^3 - 6c^2 + 8c + 2 = 0
\]
To find the roots, we can use the Rational Root Theorem to test for rational roots. Testing \(c = 2\):
\[
2^3 - 6(2^2) + 8(2) + 2 = 8 - 24 + 16 + 2 = 2 \neq 0
\]
Testing \(c = 1\):
\[
1^3 - 6(1^2) + 8(1) + 2 = 1 - 6 + 8 + 2 = 5 \neq 0
\]
Testing \(c = -1\):
\[
(-1)^3 - 6(-1)^2 + 8(-1) + 2 = -1 - 6 - 8 + 2 = -13 \neq 0
\]
Finding another root requires numerical methods or polynomial division.
Since the cubic equation does not easily yield rational roots, numerical methods or graphing may assist in finding approximate solutions. Alternatively, we could also use synthetic division or the cubic formula for exact roots.
Finally, it’s good to check for extraneous roots as well as the domain for \(c\):
1. \(c \neq 2\) and \(c \neq -2\) to avoid undefined expressions.
This shows the limitations within the rational expression.
If you'd like, I could help you with numerical approximations or a different approach to finding the roots.
\[
\frac{c-4}{c-2} = \frac{c-2}{c+2} - \frac{1}{2-c}
\]
let's first simplify the right-hand side. Notice that \(2 - c\) can be rewritten as \(-(c - 2)\). Therefore, we can rewrite the term \(\frac{1}{2 - c}\) as:
\[
-\frac{1}{c - 2}
\]
Thus, the equation becomes:
\[
\frac{c-4}{c-2} = \frac{c-2}{c+2} + \frac{1}{c - 2}
\]
Next, we need to combine the terms on the right-hand side:
\[
\text{Right side: } \frac{c-2}{c+2} + \frac{1}{c-2} = \frac{(c-2)^2 + (c+2)}{(c+2)(c-2)}
\]
The expansion of \((c-2)^2\) gives:
\[
(c-2)^2 = c^2 - 4c + 4
\]
Now add \((c + 2)\):
\[
c^2 - 4c + 4 + c + 2 = c^2 - 3c + 6
\]
So we have:
\[
\frac{c^2 - 3c + 6}{(c+2)(c-2)}
\]
Now the right-hand side becomes:
\[
\frac{c^2 - 3c + 6}{(c+2)(c-2)}
\]
Setting both sides of the equation equal:
\[
\frac{c-4}{c-2} = \frac{c^2 - 3c + 6}{(c+2)(c-2)}
\]
Now to solve this equation, we can cross-multiply:
\[
(c-4)(c+2) = (c^2 - 3c + 6)(c-2)
\]
Expanding both sides:
\[
c^2 + 2c - 4c - 8 = c^3 - 2c^2 - 3c^2 + 6c - 6
\]
This simplifies to:
\[
c^2 - 2c - 8 = c^3 - 5c^2 + 6c - 6
\]
Bringing all terms to one side gives:
\[
0 = c^3 - 5c^2 + 6c - 6 - c^2 + 2c + 8
\]
Combining like terms leads to:
\[
0 = c^3 - 6c^2 + 8c + 2
\]
Rearranging the equation:
\[
c^3 - 6c^2 + 8c + 2 = 0
\]
To find the roots, we can use the Rational Root Theorem to test for rational roots. Testing \(c = 2\):
\[
2^3 - 6(2^2) + 8(2) + 2 = 8 - 24 + 16 + 2 = 2 \neq 0
\]
Testing \(c = 1\):
\[
1^3 - 6(1^2) + 8(1) + 2 = 1 - 6 + 8 + 2 = 5 \neq 0
\]
Testing \(c = -1\):
\[
(-1)^3 - 6(-1)^2 + 8(-1) + 2 = -1 - 6 - 8 + 2 = -13 \neq 0
\]
Finding another root requires numerical methods or polynomial division.
Since the cubic equation does not easily yield rational roots, numerical methods or graphing may assist in finding approximate solutions. Alternatively, we could also use synthetic division or the cubic formula for exact roots.
Finally, it’s good to check for extraneous roots as well as the domain for \(c\):
1. \(c \neq 2\) and \(c \neq -2\) to avoid undefined expressions.
This shows the limitations within the rational expression.
If you'd like, I could help you with numerical approximations or a different approach to finding the roots.