Solve: X^2+3x+7 = 6/(x^2+3x+2)

Hey math whizzes! Ever stumbled upon an equation that looks like a tangled mess and wondered where to even begin? Today, we're diving deep into a super interesting problem: solving the quadratic equation x^2+3x+7=rac{6}{x^2+3 x+2}. This isn't your average, run-of-the-mill quadratic. It's got a bit of a twist, making it a fantastic challenge for anyone looking to sharpen their algebra skills. We'll break it down piece by piece, making sure you understand every single step. So, grab your calculators, a comfy seat, and let's get this math party started!

Understanding the Problem: What's Going On Here?

Alright guys, let's first get a solid grip on what we're dealing with. The equation we need to solve is x^2+3x+7=rac{6}{x^2+3 x+2}. At first glance, it might seem a bit intimidating because you have a variable expression in the denominator. This means we need to be extra careful about any values of 'x' that might make that denominator zero, as division by zero is a big no-no in math. The core of this problem lies in recognizing that the expression $x^2+3x$ appears in both the numerator side and within the denominator on the right side. This is our golden ticket, the key to simplifying this whole thing down into something much more manageable. Whenever you see a repeating or similar expression in a tricky equation like this, your first instinct should be to make a substitution. It's like giving a complicated character a nickname so you can refer to them more easily. This strategy will transform our complex equation into a much simpler form, which we can then solve using our trusty quadratic formula or factoring techniques. Remember, the goal is to turn that beast of an equation into a friendly, familiar form. We'll be looking for values of 'x' that make the entire equation true, and by using substitution, we can pave the way to finding those solutions efficiently and without too much head-scratching. Keep an eye out for those repeated patterns – they're usually hints from the universe telling you to simplify!

The Power of Substitution: Making it Easier

So, how do we actually use that observation that $x^2+3x$ is our repeating hero? The magic word here is substitution. We're going to introduce a new variable to represent that common part. Let's say we let $y = x^2+3x$. Now, let's look at our original equation again: x^2+3x+7=rac{6}{x^2+3 x+2}. If we substitute 'y' into this, we get a much cleaner equation. The left side, $x^2+3x+7$, becomes $y+7$. And the right side, rac{6}{x^2+3 x+2}, becomes rac{6}{y+2}. So, our intimidating equation transforms into the much simpler y+7 = rac{6}{y+2}. This is a huge step forward, guys! We've gone from a complex rational equation involving quadratics to a much simpler equation involving just one variable, 'y'. This substitution technique is a lifesaver in algebra, allowing us to break down complex problems into smaller, more digestible parts. It's like using a secret code to simplify communication. Before we proceed, we must also remember the condition we noted earlier: the denominator cannot be zero. In our original equation, $x^2+3x+2 eq 0$. In terms of our new variable 'y', this means $y+2 eq 0$, or $y eq -2$. This is a crucial constraint we'll need to check our final 'y' solutions against. If we get $y = -2$, it's an extraneous solution and must be discarded. Now that we have our simplified equation y+7 = rac{6}{y+2}, we can proceed to solve for 'y'. This is where we start getting closer to finding our actual 'x' values!

Solving for 'y': A Simple Quadratic Emerges

Now that we've got the equation in terms of 'y', y+7 = rac{6}{y+2}, we can solve for 'y'. Remember our condition: $y eq -2$. The first step is to get rid of that fraction. We do this by multiplying both sides of the equation by the denominator, $(y+2)$. So, we have: $(y+7)(y+2) = 6$. Now, let's expand the left side using the distributive property (or FOIL method if you prefer): $y imes y + y imes 2 + 7 imes y + 7 imes 2 = 6$. This simplifies to $y^2 + 2y + 7y + 14 = 6$. Combine the like terms ($2y$ and $7y$) to get $y^2 + 9y + 14 = 6$. Our goal is to set this equation to zero to solve it like a standard quadratic equation. So, subtract 6 from both sides: $y^2 + 9y + 14 - 6 = 0$. This gives us our beautiful, simple quadratic equation: $y^2 + 9y + 8 = 0$. See? That substitution really did wonders! Now we have a quadratic equation in 'y' that we can solve using factoring or the quadratic formula. Factoring is usually quicker if it's possible. We need two numbers that multiply to 8 and add up to 9. Those numbers are 1 and 8! So, we can factor the equation as $(y+1)(y+8) = 0$. For this product to be zero, at least one of the factors must be zero. This means either $y+1 = 0$ or $y+8 = 0$. Solving these gives us our two possible values for 'y': $y = -1$ and $y = -8$. Now, let's quickly check these against our condition $y eq -2$. Both $y = -1$ and $y = -8$ satisfy this condition, so they are both valid solutions for 'y'. We're almost there, guys! We've found the values of 'y', and now we just need to convert them back to find our original 'x' values.

Back to 'x': Finding the Final Solutions

We've successfully found our values for 'y': $y = -1$ and $y = -8$. But remember, the original question was to solve for 'x', not 'y'. So, it's time to reverse our substitution. We defined $y = x^2+3x$. Now, we'll set each of our 'y' values back into this equation and solve for 'x'.

Case 1: $y = -1$

Substitute $y = -1$ into $y = x^2+3x$: $-1 = x^2+3x$ Rearrange this into a standard quadratic form by adding 1 to both sides: $x^2+3x+1 = 0$

This quadratic doesn't look easily factorable, so we'll use the quadratic formula, which is x = rac{-b ypeof{ ext{±}} ext{b}^2-4ac}{2a}. In this equation, $a=1$, $b=3$, and $c=1$. Plugging these values in:

x = rac{-3 ypeof{ ext{±}} ext{3}^2-4(1)(1)}{2(1)} x = rac{-3 ypeof{ ext{±}} ext{9-4}}{2} x = rac{-3 ypeof{ ext{±}} ext{5}}{2}

This gives us two solutions for 'x': x_1 = rac{-3 + ext{√}5}{2} x_2 = rac{-3 - ext{√}5}{2}

Case 2: $y = -8$

Now, substitute $y = -8$ into $y = x^2+3x$: $-8 = x^2+3x$ Rearrange this into standard quadratic form by adding 8 to both sides: $x^2+3x+8 = 0$

Again, this quadratic doesn't look easily factorable, so we'll use the quadratic formula with $a=1$, $b=3$, and $c=8$.

x = rac{-3 ypeof{ ext{±}} ext{3}^2-4(1)(8)}{2(1)} x = rac{-3 ypeof{ ext{±}} ext{9-32}}{2} x = rac{-3 ypeof{ ext{±}} ext{-23}}{2}

Now, we have a negative number under the square root ($ ext{√} ext{-23}$). In the realm of real numbers, we cannot take the square root of a negative number. This means that for the case $y = -8$, there are no real solutions for 'x'. If we were working with complex numbers, we'd have solutions involving the imaginary unit 'i', but typically, when solving equations like this without further specification, we assume we're looking for real solutions.

Checking Our Work: The Final Verdict

So, we've arrived at our potential solutions. From Case 1 ($y = -1$), we got two real solutions for 'x': x = rac{-3 ypeof{ ext{±}} ext{√}5}{2}. From Case 2 ($y = -8$), we found no real solutions. It's always a smart move, especially in these kinds of problems, to quickly check if our solutions cause any issues with the original denominator, $x^2+3x+2$. We established that $x^2+3x+2 eq 0$. Let's see what happens when we substitute our 'y' values back.

Remember $y = x^2+3x$. So, $x^2+3x+2 = y+2$. Our condition $x^2+3x+2 eq 0$ becomes $y+2 eq 0$, which means $y eq -2$. We already checked our 'y' values ($y=-1$ and $y=-8$) against this, and neither of them is $-2$. This confirms that our 'y' values are valid and won't lead to division by zero in the original equation.

Therefore, the real solutions to the equation x^2+3x+7=rac{6}{x^2+3 x+2} are:

x = rac{-3 + ext{√}5}{2} and x = rac{-3 - ext{√}5}{2}

And that's a wrap, guys! We tackled a tricky equation by using substitution to simplify it into a standard quadratic, solved for our intermediate variable, and then worked our way back to find the original solutions. It's all about breaking down the problem and using the right tools at the right time. Keep practicing, and you'll become a math-solving ninja in no time!