Answer :
To solve the equation \(\frac{x^2}{x-2} + 5 = \frac{5x - 6}{x-2}\) and find the possible values of \(x\), let's work through it step by step.
First, we will look at the given equation:
[tex]\[ \frac{x^2}{x-2} + 5 = \frac{5x - 6}{x-2} \][/tex]
1. Combine the fractions:
Since both fractions have the same denominator, we can combine them:
[tex]\[ \frac{x^2}{x-2} + 5 = \frac{5x - 6}{x-2} \][/tex]
2. Move all terms to one side of the equation:
Subtract \(\frac{5x - 6}{x-2}\) from both sides to set the equation to zero:
[tex]\[ \frac{x^2}{x-2} + 5 - \frac{5x - 6}{x-2} = 0 \][/tex]
3. Combine the fractions under a common denominator:
Since both terms are over the same denominator \(x-2\), we can combine them into a single fraction:
[tex]\[ \frac{x^2 - (5x - 6) + 5(x - 2)}{x-2} = 0 \][/tex]
4. Simplify the numerator:
Distribute and combine like terms in the numerator:
[tex]\[ x^2 - 5x + 6 + 5x - 10 = x^2 - 4 \][/tex]
5. Write the simplified fraction:
So the equation simplifies to:
[tex]\[ \frac{x^2 - 4}{x-2} = 0 \][/tex]
6. Solve the simplified numerator:
Since a fraction is zero when its numerator is zero (and its denominator is not zero):
[tex]\[ x^2 - 4 = 0 \][/tex]
7. Factor the numerator:
The quadratic \(x^2 - 4\) can be factored using difference of squares:
[tex]\[ (x - 2)(x + 2) = 0 \][/tex]
8. Find the roots of the equation:
Set each factor to zero and solve for \(x\):
[tex]\[ x - 2 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
[tex]\[ x = 2 \quad \text{or} \quad x = -2 \][/tex]
9. Check for extraneous solutions:
- Note that the original denominator \(x-2\) must not be zero.
- \(x=2\) would make the denominator zero, so it is an extraneous solution and must be discarded.
Given the steps above, we find that the only valid solution to the equation is:
[tex]\[ x = -2 \][/tex]
So, the correct answer is:
c) -2
First, we will look at the given equation:
[tex]\[ \frac{x^2}{x-2} + 5 = \frac{5x - 6}{x-2} \][/tex]
1. Combine the fractions:
Since both fractions have the same denominator, we can combine them:
[tex]\[ \frac{x^2}{x-2} + 5 = \frac{5x - 6}{x-2} \][/tex]
2. Move all terms to one side of the equation:
Subtract \(\frac{5x - 6}{x-2}\) from both sides to set the equation to zero:
[tex]\[ \frac{x^2}{x-2} + 5 - \frac{5x - 6}{x-2} = 0 \][/tex]
3. Combine the fractions under a common denominator:
Since both terms are over the same denominator \(x-2\), we can combine them into a single fraction:
[tex]\[ \frac{x^2 - (5x - 6) + 5(x - 2)}{x-2} = 0 \][/tex]
4. Simplify the numerator:
Distribute and combine like terms in the numerator:
[tex]\[ x^2 - 5x + 6 + 5x - 10 = x^2 - 4 \][/tex]
5. Write the simplified fraction:
So the equation simplifies to:
[tex]\[ \frac{x^2 - 4}{x-2} = 0 \][/tex]
6. Solve the simplified numerator:
Since a fraction is zero when its numerator is zero (and its denominator is not zero):
[tex]\[ x^2 - 4 = 0 \][/tex]
7. Factor the numerator:
The quadratic \(x^2 - 4\) can be factored using difference of squares:
[tex]\[ (x - 2)(x + 2) = 0 \][/tex]
8. Find the roots of the equation:
Set each factor to zero and solve for \(x\):
[tex]\[ x - 2 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
[tex]\[ x = 2 \quad \text{or} \quad x = -2 \][/tex]
9. Check for extraneous solutions:
- Note that the original denominator \(x-2\) must not be zero.
- \(x=2\) would make the denominator zero, so it is an extraneous solution and must be discarded.
Given the steps above, we find that the only valid solution to the equation is:
[tex]\[ x = -2 \][/tex]
So, the correct answer is:
c) -2
