Equation Solving: Spotting Extraneous Solutions

Hey math whizzes! Today, we're diving deep into the wild world of equations, specifically tackling one that's a bit of a tricky beast: 1+rac{3}{c^2-11 c+30}=rac{3}{c-6}. Now, this isn't just about finding 'c'; it's a full-on detective mission where we need to check for extraneous solutions. You know, those sneaky values that seem to work at first but turn out to be imposters. Let's break this down, step by step, and make sure we nail it!

Understanding Extraneous Solutions: What's the Deal?

Before we even touch our equation, let's get our heads around what extraneous solutions are. Think of them as the 'fake news' of the math world. When we manipulate equations, especially those with variables in the denominator (like ours here, with 'c' hanging out in $c^2-11 c+30$ and $c-6$), we sometimes introduce solutions that don't actually satisfy the original equation. This usually happens when we multiply both sides of an equation by an expression that could be zero. If that expression is zero for a particular value of our variable, then that value can become an extraneous solution. The golden rule here, guys, is always check your answers by plugging them back into the original equation. It's the ultimate truth serum for our solutions. If it makes any denominator zero, poof, it's extraneous!

Factoring the Denominator: Our First Big Move

Our equation is 1+rac{3}{c^2-11 c+30}=rac{3}{c-6}. The first thing that should scream at you is that quadratic expression in the denominator: $c^2-11 c+30$. To make this whole thing easier to handle, we need to factor it. We're looking for two numbers that multiply to 30 and add up to -11. After a little brain-wracking, we find those magic numbers are -5 and -6. So, $c^2-11 c+30$ factors into $(c-5)(c-6)$. This is a huge step because it reveals the common factors and potential pitfalls. Now our equation looks like this: 1+rac{3}{(c-5)(c-6)}=rac{3}{c-6}. See that $(c-6)$ in both denominators? That's our clue!

Identifying Potential Issues: The Forbidden Values

With the denominator factored, we can now clearly see which values of 'c' would cause trouble. Remember, we can never divide by zero. So, for our equation 1+rac{3}{(c-5)(c-6)}=rac{3}{c-6}, the denominators cannot be zero. This means:

• $c-5 eq 0 ightarrow c eq 5$
• $c-6 eq 0 ightarrow c eq 6$

These values, $c=5$ and $c=6$, are our forbidden values. If we find any solutions that end up being 5 or 6, we know immediately they are extraneous. Keep these in your back pocket, because we'll be using them later to filter our results.

Clearing the Denominators: The Least Common Multiple (LCM)

To solve for 'c', we need to get rid of those pesky denominators. The best way to do this is to multiply every single term in the equation by the Least Common Multiple (LCM) of the denominators. Our denominators are $(c-5)(c-6)$ and $(c-6)$. The LCM of these is clearly $(c-5)(c-6)$. So, let's multiply through:

(c-5)(c-6) imes 1 + (c-5)(c-6) imes rac{3}{(c-5)(c-6)} = (c-5)(c-6) imes rac{3}{c-6}

Now, let's simplify term by term:

• The first term is easy: $(c-5)(c-6) imes 1 = (c-5)(c-6)$
• The second term: The $(c-5)$ and $(c-6)$ in the numerator cancel out the $(c-5)(c-6)$ in the denominator, leaving us with just 3.
• The third term: The $(c-6)$ in the numerator cancels out the $(c-6)$ in the denominator, leaving us with $(c-5) imes 3$, which is $3(c-5)$.

So, our equation simplifies to: $(c-5)(c-6) + 3 = 3(c-5)$.

Solving the Simplified Equation: Finding Our Candidates

We're getting closer, guys! Now we have a much simpler equation to solve: $(c-5)(c-6) + 3 = 3(c-5)$. Let's expand and rearrange this to find our potential solutions.

First, expand the products:

• $(c-5)(c-6) = c^2 - 6c - 5c + 30 = c^2 - 11c + 30$
• $3(c-5) = 3c - 15$

Substitute these back into our equation:

$c^2 - 11c + 30 + 3 = 3c - 15$

Combine like terms on the left side:

$c^2 - 11c + 33 = 3c - 15$

Now, let's move all terms to one side to set the equation to zero, aiming for that standard quadratic form ($ax^2 + bx + c = 0$):

$c^2 - 11c - 3c + 33 + 15 = 0$

$c^2 - 14c + 48 = 0$

We've got ourselves a quadratic equation! Now we need to find the values of 'c' that satisfy this. We can try factoring again. We're looking for two numbers that multiply to 48 and add up to -14. Let's think... -6 and -8 work perfectly! They multiply to 48 and add to -14.

So, we can factor the quadratic as:

$(c-6)(c-8) = 0$

This equation is true if either factor is zero:

• $c-6 = 0 ightarrow c = 6$
• $c-8 = 0 ightarrow c = 8$

So, our potential solutions are $c=6$ and $c=8$. But wait! Remember our detective work earlier? We identified some forbidden values.

The Grand Finale: Checking for Extraneous Solutions

This is the most crucial step, folks! We found two potential solutions, $c=6$ and $c=8$, but we must check them against our original equation and our forbidden values. Remember those values we identified that would make a denominator zero? They were $c eq 5$ and $c eq 6$.

Let's examine our first potential solution: $c=6$.

Is $c=6$ one of our forbidden values? YES! We found earlier that $c$ cannot be 6 because it would make the denominator $c-6$ (and also $(c-5)(c-6)$) equal to zero. If we were to plug $c=6$ back into the original equation 1+rac{3}{c^2-11 c+30}=rac{3}{c-6}, we'd get rac{3}{6-6} = rac{3}{0}, which is undefined. Therefore, $c=6$ is an extraneous solution.

Now, let's look at our second potential solution: $c=8$.

Is $c=8$ one of our forbidden values? No, 8 is not equal to 5 or 6. Let's plug it back into the original equation to be absolutely sure:

Original equation: 1+rac{3}{c^2-11 c+30}=rac{3}{c-6}

Substitute $c=8$:

Left side: 1+rac{3}{8^2-11(8)+30} = 1+rac{3}{64-88+30} = 1+rac{3}{64-58} = 1+rac{3}{6}

1+rac{3}{6} = 1+rac{1}{2} = rac{3}{2}

Right side: rac{3}{c-6} = rac{3}{8-6} = rac{3}{2}

Since the left side (rac{3}{2}) equals the right side (rac{3}{2}), $c=8$ is a valid solution.

Conclusion: The Verified Answer

So, after all that hard work, what's the final verdict? We started with the equation 1+rac{3}{c^2-11 c+30}=rac{3}{c-6}. We factored, cleared denominators, solved the resulting quadratic, and crucially, checked our potential answers against the original equation and the values that would cause division by zero. We found two candidates, $c=6$ and $c=8$. However, $c=6$ caused a division by zero in the original equation, making it an extraneous solution. Only $c=8$ held up under scrutiny. Therefore, the only valid solution to the equation is $c=8$. Remember, guys, the check for extraneous solutions is not optional – it's the key to solving these types of rational equations correctly! Keep practicing, and you'll become pros at spotting those fakes in no time!