Question

1. solve the rational equation \\(\frac{4}{k^2 - 8k + 12} = \frac{k}{k - 2} + \frac{1}{k - 6}\\). identify any non - permissible values and verify the solution(s).

Explanation:

Step1: Factor the denominator

Factor \(k^{2}-8k + 12\): \(k^{2}-8k + 12=(k - 2)(k - 6)\)
So the equation becomes \(\frac{4}{(k - 2)(k - 6)}=\frac{k}{k - 2}+\frac{1}{k - 6}\)
Non - permissible values: The denominators cannot be zero. So \(k-2
eq0\) (i.e., \(k
eq2\)) and \(k - 6
eq0\) (i.e., \(k
eq6\))

Step2: Eliminate the denominators

Multiply each term by the least common denominator \((k - 2)(k - 6)\) to get rid of the fractions:
\(4=k(k - 6)+1\times(k - 2)\)

Step3: Expand and simplify the right - hand side

Expand \(k(k - 6)+(k - 2)\):
\(k(k - 6)=k^{2}-6k\) and \(k - 2\) remains as is.
So \(k^{2}-6k + k-2=k^{2}-5k - 2\)
The equation is now \(4=k^{2}-5k - 2\)

Step4: Rearrange the equation to standard quadratic form

Subtract 4 from both sides: \(k^{2}-5k-2 - 4=0\)
Simplify to get \(k^{2}-5k - 6=0\)

Step5: Solve the quadratic equation

Factor the quadratic equation \(k^{2}-5k - 6\):
We need two numbers that multiply to - 6 and add to - 5. The numbers are - 6 and 1.
So \(k^{2}-5k - 6=(k - 6)(k+1)=0\)
Set each factor equal to zero:
\(k - 6=0\) gives \(k = 6\) (but \(k = 6\) is a non - permissible value, so we discard it)
\(k + 1=0\) gives \(k=-1\)

Step6: Verify the solution

Substitute \(k = - 1\) into the original equation:
Left - hand side: \(\frac{4}{(-1)^{2}-8\times(-1)+12}=\frac{4}{1 + 8+12}=\frac{4}{21}\)
Right - hand side: \(\frac{-1}{-1 - 2}+\frac{1}{-1 - 6}=\frac{-1}{-3}+\frac{1}{-7}=\frac{1}{3}-\frac{1}{7}=\frac{7 - 3}{21}=\frac{4}{21}\)
Since the left - hand side equals the right - hand side when \(k=-1\), \(k = - 1\) is a valid solution.

Answer:

Non - permissible values: \(k
eq2\) and \(k
eq6\); Solution: \(k=-1\)