Solving $x+4=\sqrt{8x+25}$ For $x$

When you're faced with an equation like $x+4=\sqrt{8x+25}$, it might look a little intimidating at first glance. The presence of a square root often means we need to be a bit clever in how we approach it. But don't worry, with a few strategic steps, we can unravel this mathematical mystery and find the value(s) of $x$ that make this equation true. Our main goal here is to isolate $x$, and the first hurdle is that pesky square root. So, how do we get rid of it? The most common and effective method is to square both sides of the equation. This is a powerful technique because $(\sqrt{a})^2 = a$, which effectively removes the radical. However, we must be mindful that squaring both sides can sometimes introduce extraneous solutions. These are solutions that arise from the squaring process but don't actually satisfy the original equation. Therefore, it's absolutely crucial that we check our final answers in the original equation to ensure they are valid.

Before we square, it's also a good idea to consider the domain of the equation. The expression under the square root, $8x+25$, must be non-negative. This means $8x+25 \ge 0$, which simplifies to $8x \ge -25$, or $x \ge -25/8$. Additionally, since the square root symbol ($\sqrt{}$) denotes the principal (non-negative) square root, the left side of the equation, $x+4$, must also be non-negative. This implies $x+4 \ge 0$, which means $x \ge -4$. Combining these conditions, our potential solutions must satisfy $x \ge -4$. This initial check helps us eliminate any results that are impossible from the outset.

Now, let's get down to the business of solving. We start with our equation: $x+4 = \sqrt{8x+25}$. To eliminate the square root, we square both sides: $(x+4)^2 = (\sqrt{8x+25})^2$. On the left side, we expand $(x+4)^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$. So, $(x+4)^2 = x^2 + 2(x)(4) + 4^2 = x^2 + 8x + 16$. On the right side, the square root and the square cancel each other out, leaving us with $8x+25$. Now our equation looks like this: $x^2 + 8x + 16 = 8x + 25$. This is a quadratic equation, and our next step is to rearrange it into the standard form $ax^2 + bx + c = 0$. To do this, we move all terms to one side. Let's subtract $8x$ from both sides and subtract $25$ from both sides:

$x^2 + 8x + 16 - 8x - 25 = 0$

Combining like terms, the $8x$ terms cancel out ($8x - 8x = 0$), and we are left with:

$x^2 + (16 - 25) = 0$

$x^2 - 9 = 0$

This is a very simple quadratic equation! We can solve it in a couple of ways. One way is to add 9 to both sides to get $x^2 = 9$. Then, we take the square root of both sides. Remember that when you take the square root of a number, there are two possible solutions: a positive one and a negative one. So, $x = \pm\sqrt{9}$. This gives us two potential solutions: $x = 3$ and $x = -3$.

Another way to solve $x^2 - 9 = 0$ is by factoring it as a difference of squares, since $x^2$ is the square of $x$ and $9$ is the square of $3$. The difference of squares formula is $a^2 - b^2 = (a-b)(a+b)$. Applying this here, we get $(x-3)(x+3) = 0$. For this product to be zero, at least one of the factors must be zero. So, either $x-3 = 0$ (which gives $x = 3$) or $x+3 = 0$ (which gives $x = -3$). Both methods yield the same potential solutions: $x=3$ and $x=-3$.

Checking for Extraneous Solutions

As we mentioned earlier, squaring both sides of an equation can sometimes introduce extraneous solutions. Therefore, it is absolutely essential that we check our potential solutions, $x=3$ and $x=-3$, in the original equation: $x+4 = \sqrt{8x+25}$. Let's start with $x=3$. Substitute $3$ for $x$ in the original equation:

Left side: $x+4 = 3+4 = 7$

Right side: $\sqrt{8x+25} = \sqrt{8(3)+25} = \sqrt{24+25} = \sqrt{49} = 7$

Since the left side ($7$) equals the right side ($7$), $x=3$ is a valid solution. It satisfies the original equation.

Now, let's check the second potential solution, $x=-3$. Substitute $-3$ for $x$ in the original equation:

Left side: $x+4 = -3+4 = 1$

Right side: $\sqrt{8x+25} = \sqrt{8(-3)+25} = \sqrt{-24+25} = \sqrt{1} = 1$

Since the left side ($1$) equals the right side ($1$), $x=-3$ is also a valid solution. It satisfies the original equation.

Wait, let's re-evaluate our initial domain restrictions. We found that for the square root to be defined and for the equation to hold, we needed $x \ge -4$. Both $x=3$ and $x=-3$ satisfy this condition ($3 \ge -4$ and $-3 \ge -4$).

However, there's a subtle point we need to re-examine carefully. Remember the condition that $x+4$ must be non-negative because it's equal to a principal square root? This means $x+4 ge 0$, or $x ge -4$. Let's re-check our solutions against this specific condition.

For $x=3$: $3+4 = 7$, which is indeed $\ge 0$. So $x=3$ is valid.

For $x=-3$: $-3+4 = 1$, which is also $\ge 0$. So $x=-3$ is also valid.

It seems our initial check was thorough, and both solutions hold up. However, there was a slight oversight in the initial domain reasoning which I need to correct to avoid confusion. Let's revisit the squaring step and its implications.

When we square both sides of an equation, we are essentially solving $(x+4)^2 = 8x+25$. This quadratic equation $x^2-9=0$ has solutions $x=3$ and $x=-3$. These are the values of $x$ that make the squared equation true. The original equation is $x+4=\sqrt{8x+25}$.

Let's think about what happens if we had, hypothetically, an equation like $x+4 = -\sqrt{8x+25}$. If we squared both sides here, we would get $(x+4)^2 = (-\sqrt{8x+25})^2$, which simplifies to $(x+4)^2 = 8x+25$. This is the exact same quadratic equation we solved! This is why extraneous solutions can arise. A solution to the squared equation might satisfy $x+4 = -\sqrt{8x+25}$ instead of the original $x+4 = \sqrt{8x+25}$.

So, the critical check is to ensure that $x+4$ is non-negative because it's equated to the principal square root, which is always non-negative.

Let's re-check with $x=3$: $x+4 = 3+4 = 7$. $\sqrt{8x+25} = \sqrt{8(3)+25} = \sqrt{49} = 7$. $7 = 7$. Valid.

Let's re-check with $x=-3$: $x+4 = -3+4 = 1$. $\sqrt{8x+25} = \sqrt{8(-3)+25} = \sqrt{1} = 1$. $1 = 1$. Valid.

It appears both solutions are indeed valid. My apologies for the slight confusion in the domain explanation earlier; the core principle of checking in the original equation is the most robust method, and in this case, both $x=3$ and $x=-3$ satisfy it. The condition $x ge -4$ is implicitly satisfied by the process of checking in the original equation where $x+4$ must equal a non-negative value.

Final Solution

After carefully squaring both sides, rearranging into a quadratic equation, solving the quadratic equation, and most importantly, checking both potential solutions in the original equation, we find that both $x=3$ and $x=-3$ are valid solutions. The process of solving radical equations often requires this diligent checking to ensure accuracy.

If you're looking to deepen your understanding of algebraic equations, exploring resources on solving radical equations can be very beneficial. You might find the detailed explanations and examples on websites like Khan Academy to be incredibly helpful in mastering these concepts. Their comprehensive approach covers various types of equations and provides ample practice opportunities.