Solve $5x = \sqrt{7x+5}$

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics to tackle a tricky equation: $5x = \sqrt{7x+5}$. This might look a bit intimidating at first glance, but don't worry, we'll break it down step-by-step. Our goal is to find the values of x that make this equation true. When you're dealing with equations involving square roots, the key is often to isolate the square root term and then square both sides to get rid of it. However, you've got to be super careful because squaring both sides can sometimes introduce extraneous solutions, which are solutions that work in the squared equation but not in the original one. So, after we find our potential solutions, it's absolutely crucial to plug them back into the original equation to verify they are correct. This little check is your safety net to ensure you've got the real deal. We'll explore the algebraic manipulations needed and discuss why each step is important. Get ready to flex those math muscles!

Understanding the Problem and Initial Steps

Alright, let's get started with our equation: $5x = \sqrt{7x+5}$. The first thing we notice is the square root on the right side. To make progress, we want to eliminate this square root. The standard way to do this is by squaring both sides of the equation. But before we do that, let's think about the implications. The square root symbol, $\sqrt{\cdot}$, by definition, represents the principal (non-negative) square root. This means that the right side of our equation, $\sqrt{7x+5}$, must be greater than or equal to zero. Consequently, the left side, $5x$, must also be greater than or equal to zero. This immediately tells us that any valid solution x must be non-negative, meaning $x \ge 0$. This is a super important condition to keep in mind as we proceed, and it will help us filter out any extraneous solutions later on. Now, let's square both sides of the equation:

$(5x)^2 = (\sqrt{7x+5})^2$

This simplifies to:

$25x^2 = 7x+5$

Now, we've transformed our original equation with a square root into a standard quadratic equation. Our next step is to rearrange this into the familiar $ax^2 + bx + c = 0$ form so we can solve for x.

Solving the Quadratic Equation

We have the equation $25x^2 = 7x+5$. To put it in the standard quadratic form, we need to move all terms to one side. Let's subtract $7x$ and $5$ from both sides:

$25x^2 - 7x - 5 = 0$

This is a quadratic equation where $a = 25$, $b = -7$, and $c = -5$. Now, we can use the quadratic formula to find the values of x. The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's plug in our values for a, b, and c:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(25)(-5)}}{2(25)}$

$x = \frac{7 \pm \sqrt{49 - (-500)}}{50}$

$x = \frac{7 \pm \sqrt{49 + 500}}{50}$

$x = \frac{7 \pm \sqrt{549}}{50}$

Now, we need to simplify the square root of 549 if possible. Let's find the prime factorization of 549. We can see that 549 is divisible by 3 (since the sum of its digits, $5+4+9=18$, is divisible by 3). $549 / 3 = 183$. And 183 is also divisible by 3 ($1+8+3=12$). $183 / 3 = 61$. 61 is a prime number. So, $549 = 3 \times 3 \times 61 = 9 \times 61$.

Therefore, $\sqrt{549} = \sqrt{9 \times 61} = \sqrt{9} \times \sqrt{61} = 3\sqrt{61}$.

Substituting this back into our formula for x:

$x = \frac{7 \pm 3\sqrt{61}}{50}$

This gives us two potential solutions:

1. $x_1 = \frac{7 + 3\sqrt{61}}{50}$
2. $x_2 = \frac{7 - 3\sqrt{61}}{50}$

These are the solutions we obtained after squaring both sides. But remember our initial condition? We said that x must be non-negative ($x \ge 0$). Let's examine these two potential solutions.

Checking for Extraneous Solutions

This is the critical part, guys! We need to check if our potential solutions satisfy the original equation $5x = \sqrt{7x+5}$ and the condition $x \ge 0$. Let's analyze our two potential solutions:

Solution 1: $x_1 = \frac{7 + 3\sqrt{61}}{50}$

First, let's check if $x_1 \ge 0$. Since $\sqrt{61}$ is a positive number (approximately 7.8), $3\sqrt{61}$ is positive. Adding 7 to a positive number results in a positive number. Dividing by 50 (a positive number) keeps it positive. So, $x_1$ is definitely greater than 0. This looks promising!

Now, let's plug it back into the original equation. This can be a bit tedious, but it's the only way to be sure.

Left side: $5x_1 = 5 \left(\frac{7 + 3\sqrt{61}}{50}\right) = \frac{7 + 3\sqrt{61}}{10}$

Right side: $\sqrt{7x_1+5} = \sqrt{7\left(\frac{7 + 3\sqrt{61}}{50}\right) + 5}$

$= \sqrt{\frac{49 + 21\sqrt{61}}{50} + \frac{250}{50}}$

$= \sqrt{\frac{299 + 21\sqrt{61}}{50}}$

To see if the left side equals the right side, we can square the left side and see if it matches what's under the square root on the right side. This is equivalent to checking if $25x_1^2 = 7x_1+5$, which we already know is true from how we derived $x_1$. However, a more direct check is to see if $\left(\frac{7 + 3\sqrt{61}}{10}\right)^2$ equals $\frac{299 + 21\sqrt{61}}{50}$.

$\left(\frac{7 + 3\sqrt{61}}{10}\right)^2 = \frac{(7 + 3\sqrt{61})^2}{100} = \frac{7^2 + 2(7)(3\sqrt{61}) + (3\sqrt{61})^2}{100}$

$= \frac{49 + 42\sqrt{61} + 9(61)}{100} = \frac{49 + 42\sqrt{61} + 549}{100} = \frac{598 + 42\sqrt{61}}{100}$

This doesn't immediately look like $\frac{299 + 21\sqrt{61}}{50}$. Let's simplify the fraction by dividing the numerator and denominator by 2:

$\frac{598 + 42\sqrt{61}}{100} = \frac{299 + 21\sqrt{61}}{50}$

And indeed, it matches! So, $x_1 = \frac{7 + 3\sqrt{61}}{50}$ is a valid solution. The option A given in the problem statement is $\frac{7+3 \sqrt{61}}{50}$, which matches our first solution.

Solution 2: $x_2 = \frac{7 - 3\sqrt{61}}{50}$

Now, let's check our second potential solution. We need to see if $x_2 \ge 0$. We know $\sqrt{61}$ is approximately 7.8. So, $3\sqrt{61}$ is approximately $3 \times 7.8 = 23.4$. Then, $7 - 3\sqrt{61}$ is approximately $7 - 23.4 = -16.4$. Since this value is negative, dividing by 50 will also result in a negative number. Therefore, $x_2 = \frac{7 - 3\sqrt{61}}{50} < 0$.

This violates our initial condition that $x \ge 0$. If we were to plug this value back into the original equation, the left side $5x_2$ would be negative, while the right side $\sqrt{7x_2+5}$ must be non-negative by definition of the square root. Thus, this solution is extraneous.

What about the second option given, $\frac{7+i \sqrt{451}}{50}$? The presence of the imaginary unit 'i' indicates a complex number. Our original equation involves real numbers and the principal square root, which typically operates within the domain of real numbers unless specified otherwise. The steps we took led us to real number solutions. If the discriminant ($b^2-4ac$) were negative, we would get complex solutions from the quadratic formula. In our case, the discriminant was $549$, which is positive, leading to real solutions. The option B seems to arise from a misunderstanding or a different problem altogether, as our derived solutions are real and one of them is extraneous.

Conclusion

After carefully solving the quadratic equation derived from $5x = \sqrt{7x+5}$ and rigorously checking for extraneous solutions, we found that only one solution satisfies the original equation and its inherent constraints. The potential solutions were $x = \frac{7 \pm 3\sqrt{61}}{50}$. We determined that $x_1 = \frac{7 + 3\sqrt{61}}{50}$ is a valid solution because it is positive and satisfies the original equation. On the other hand, $x_2 = \frac{7 - 3\sqrt{61}}{50}$ is an extraneous solution because it is negative, which contradicts the requirement that $5x$ must be non-negative (as it equals a square root). Therefore, the only correct solution to the equation $5x = \sqrt{7x+5}$ is $x = \frac{7 + 3\sqrt{61}}{50}$. This matches option A. The presence of complex numbers in option B indicates it's not a solution to this particular problem. Keep practicing these checks, guys, they're super important in algebra!