Solving Radical Equations An Example With $\sqrt{7y^2 + 15y} - 2y = 5 + Y$

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In this article, we will explore the step-by-step process of solving the equation $\sqrt{7y^2 + 15y} - 2y = 5 + y$. This equation involves a square root, which requires careful manipulation to isolate the variable y. Our goal is to find all possible solutions and verify them to ensure they are valid. We will cover each step in detail, explaining the reasoning behind the operations and highlighting potential pitfalls. Understanding how to solve such equations is crucial in various areas of mathematics and its applications. Let's dive in and find the solutions to this equation.

Step 1: Isolate the Square Root

The first crucial step in solving an equation with a square root is to isolate the square root term on one side of the equation. This makes it easier to eliminate the square root by squaring both sides in the subsequent steps. By isolating the square root, we set the stage for simplifying the equation and eventually solving for the variable. The initial equation we are working with is:

$\sqrt{7y^2 + 15y} - 2y = 5 + y$

To isolate the square root term, $\sqrt{7y^2 + 15y}$, we need to move the $-2y$ term to the right side of the equation. We can do this by adding $2y$ to both sides. This maintains the balance of the equation and moves us closer to our goal of isolating the square root. Here’s the process:

$\sqrt{7y^2 + 15y} - 2y + 2y = 5 + y + 2y$

Simplifying both sides, we get:

$\sqrt{7y^2 + 15y} = 3y + 5$

Now, the square root term is isolated on the left side of the equation, which sets us up perfectly for the next step of squaring both sides to eliminate the square root. This is a critical transformation that simplifies the equation and allows us to work with polynomial terms. Isolating the square root is not just a procedural step; it's a strategic move that significantly simplifies the algebraic manipulations that follow.

Step 2: Square Both Sides

After isolating the square root, the next logical step is to eliminate it. We achieve this by squaring both sides of the equation. Squaring both sides removes the square root, allowing us to work with a simpler polynomial equation. However, it's essential to remember that squaring both sides can sometimes introduce extraneous solutions, which we'll need to check later. So, starting from our equation: $\sqrt{7y^2 + 15y} = 3y + 5$, we will square both sides:

$(\sqrt{7y^2 + 15y})^2 = (3y + 5)^2$

On the left side, squaring the square root simply removes the radical:

$7y^2 + 15y = (3y + 5)^2$

Now, let’s expand the right side. We need to square the binomial $(3y + 5)$. Remember that $(a + b)^2 = a^2 + 2ab + b^2$, so:

$(3y + 5)^2 = (3y)^2 + 2(3y)(5) + (5)^2$

$(3y + 5)^2 = 9y^2 + 30y + 25$

So, our equation becomes:

$7y^2 + 15y = 9y^2 + 30y + 25$

This is now a quadratic equation, which we can solve using standard algebraic techniques. Squaring both sides has transformed our equation from one involving a square root to a more manageable polynomial form. The next step will involve rearranging the terms to bring everything to one side, setting the equation to zero, and then attempting to solve for y.

Step 3: Simplify and Rearrange

Now that we have squared both sides of the equation, we have a quadratic equation to solve. The equation we derived is:

$7y^2 + 15y = 9y^2 + 30y + 25$

To solve this quadratic equation, the first step is to rearrange the terms so that all terms are on one side, and the equation is set to zero. This standard form ($ax^2 + bx + c = 0$) is essential for applying methods like factoring, completing the square, or using the quadratic formula. Let's move all terms to the right side of the equation to ensure the coefficient of the $y^2$ term is positive, which can simplify our work.

Subtract $7y^2$ and $15y$ from both sides:

$7y^2 + 15y - 7y^2 - 15y = 9y^2 + 30y + 25 - 7y^2 - 15y$

This simplifies to:

$0 = 2y^2 + 15y + 25$

Now we have a standard quadratic equation:

$2y^2 + 15y + 25 = 0$

This form allows us to proceed with methods to find the values of y that satisfy the equation. The next logical step is to attempt to factor the quadratic expression, as this is often the quickest route to finding solutions. If factoring is not straightforward, we can consider using the quadratic formula, which is a universally applicable method for solving quadratic equations. The rearrangement and simplification of the equation into this standard quadratic form is a critical step in solving for y.

Step 4: Solve the Quadratic Equation

Having rearranged the equation into the standard quadratic form $2y^2 + 15y + 25 = 0$, we now need to *solve for y. The most common methods for solving quadratic equations are factoring, completing the square, and using the quadratic formula. Factoring is often the quickest method if the quadratic expression can be easily factored. Let's first try factoring this equation. We are looking for two binomials such that:

$(ay + b)(cy + d) = 2y^2 + 15y + 25$

We need to find values for a, b, c, and d that satisfy this condition. Since the coefficient of $y^2$ is 2, we can start by considering factors of 2, which are 1 and 2. The constant term is 25, so we need to consider pairs of factors of 25, such as (1, 25) and (5, 5). By trial and error, we can find that:

$(2y + 5)(y + 5) = 2y^2 + 10y + 5y + 25 = 2y^2 + 15y + 25$

So, the quadratic equation factors as:

$(2y + 5)(y + 5) = 0$

Now, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for y:

$2y + 5 = 0$ or $y + 5 = 0$

Solving these linear equations:

For $2y + 5 = 0$:

$2y = -5$

$y = -\frac{5}{2} = -2.5$

For $y + 5 = 0$:

$y = -5$

So, we have found two potential solutions: $y = -2.5$ and $y = -5$. However, since we squared both sides of the equation earlier, we must check these solutions in the original equation to make sure they are not extraneous solutions. This is a critical step to ensure the validity of our solutions.

Step 5: Check for Extraneous Solutions

As mentioned earlier, whenever we square both sides of an equation, it is crucial to check for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original equation. These can arise because the squaring operation can introduce solutions that do not truly fit the initial problem. We found two potential solutions: $y = -2.5$ and $y = -5$. Now, we will substitute each of these values back into the original equation: $\sqrt{7y^2 + 15y} - 2y = 5 + y$ to see if they hold true.

Checking $y = -2.5$

Substitute $y = -2.5$ into the original equation:

$\sqrt{7(-2.5)^2 + 15(-2.5)} - 2(-2.5) = 5 + (-2.5)$

First, calculate the value inside the square root:

$7(-2.5)^2 + 15(-2.5) = 7(6.25) - 37.5 = 43.75 - 37.5 = 6.25$

So, the equation becomes:

$\sqrt{6.25} - 2(-2.5) = 5 + (-2.5)$

$\sqrt{6.25} = 2.5$, so:

$2.5 - (-5) = 2.5 + 5 = 7.5$

And on the right side:

$5 + (-2.5) = 2.5$

So, we have:

$7.5 = 2.5$

This is not true, so $y = -2.5$ is an extraneous solution and must be discarded.

Checking $y = -5$

Substitute $y = -5$ into the original equation:

$\sqrt{7(-5)^2 + 15(-5)} - 2(-5) = 5 + (-5)$

First, calculate the value inside the square root:

$7(-5)^2 + 15(-5) = 7(25) - 75 = 175 - 75 = 100$

So, the equation becomes:

$\sqrt{100} - 2(-5) = 5 + (-5)$

$\sqrt{100} = 10$, so:

$10 - (-10) = 10 + 10 = 20$

And on the right side:

$5 + (-5) = 0$

So, we have:

$20 = 0$

This is also not true, so $y = -5$ is an extraneous solution and must be discarded as well.

Step 6: Final Answer

After meticulously solving the equation $\sqrt{7y^2 + 15y} - 2y = 5 + y$, we have arrived at a critical juncture. We initially isolated the square root, squared both sides, simplified the resulting quadratic equation, and found two potential solutions: $y = -2.5$ and $y = -5$. However, the crucial step of checking for extraneous solutions revealed that neither of these values satisfies the original equation. This leads us to an important conclusion: The equation has no solutions. Extraneous solutions are a common occurrence when dealing with radical equations, underscoring the necessity of verifying all potential solutions in the original equation. In this case, the algebraic manipulations led us to values that appeared to be solutions but, in reality, did not hold true when subjected to the original constraints of the equation. Therefore, the final and correct answer is that there are no solutions to the given equation.

Conclusion

In this comprehensive walkthrough, we tackled the equation $\sqrt{7y^2 + 15y} - 2y = 5 + y$. We systematically isolated the square root, squared both sides, simplified the resulting quadratic equation, and arrived at potential solutions. However, we emphasized the critical importance of checking for extraneous solutions, which ultimately led us to the conclusion that the equation has no valid solutions. This process highlights the intricacies of solving radical equations and the necessity of careful verification. Understanding each step and the underlying principles is essential for success in algebra and beyond. Always remember to check your solutions in the original equation to ensure their validity and avoid the pitfall of extraneous solutions.

E. The equation has no solutions.