Solving $\sqrt{7 Y^2+15 Y}-2 Y=5+y$ A Step-by-Step Guide

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Introduction

In this article, we will delve into the step-by-step solution of the given equation: $\sqrt{7y^2 + 15y} - 2y = 5 + y$. This equation involves a square root, making it a radical equation. Solving such equations requires careful algebraic manipulation to isolate the variable y. We'll explore each step in detail, ensuring a clear understanding of the process. Our goal is to find all possible solutions and verify their correctness by substituting them back into the original equation. The solutions we find must satisfy the original equation, and we must also ensure that the expressions under the square root remain non-negative. Let's begin by isolating the square root term, which is a crucial first step in solving radical equations.

Step 1: Isolating the Square Root

The first crucial step in solving the equation $\sqrt{7y^2 + 15y} - 2y = 5 + y$ is to isolate the square root term. This involves moving all other terms to the opposite side of the equation. To achieve this, we will add 2y to both sides of the equation. This operation cancels out the -2y term on the left side, leaving the square root term by itself. By doing so, we prepare the equation for the next step, which involves squaring both sides to eliminate the square root. Isolating the square root term simplifies the subsequent algebraic manipulations and allows us to address the radical directly.

The initial equation is: $\sqrt{7y^2 + 15y} - 2y = 5 + y$.

Adding 2y to both sides gives us:

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

Now, with the square root isolated, we can proceed to the next step: squaring both sides of the equation. This will help us eliminate the square root and transform the equation into a more manageable form, typically a polynomial equation, which we can then solve using standard algebraic techniques.

Step 2: Squaring Both Sides

With the square root term isolated, the next step is to eliminate the square root by squaring both sides of the equation. This is a standard technique for solving radical equations. When we square both sides of $\sqrt{7y^2 + 15y} = 3y + 5$, we need to ensure we apply the squaring operation correctly. Squaring the left side simply removes the square root, leaving the expression inside the radical. On the right side, we have a binomial (3y + 5) which needs to be squared using the formula (a + b)^2 = a^2 + 2ab + b^2. This step is critical as it transforms the equation into a polynomial equation that can be solved using algebraic methods.

Squaring both sides of $\sqrt{7y^2 + 15y} = 3y + 5$ gives:

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

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

Expanding the right side using the formula $(a + b)^2 = a^2 + 2ab + b^2$:

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

Now, we have a quadratic equation. The next step involves rearranging the terms to bring all terms to one side, setting the equation equal to zero. This standard form allows us to use various methods, such as factoring or the quadratic formula, to find the solutions for y.

Step 3: Simplifying to a Quadratic Equation

After squaring both sides, we obtained the equation $7y^2 + 15y = 9y^2 + 30y + 25$. To solve for y, we need to rearrange this into a standard quadratic equation form, which is $ay^2 + by + c = 0$. This involves moving all terms to one side of the equation, resulting in a quadratic equation that we can then attempt to solve. Rearranging into standard form is crucial because it allows us to easily identify the coefficients a, b, and c, which are needed for both factoring and applying the quadratic formula.

To rearrange, we subtract $7y^2$ and $15y$ from both sides:

$0 = 9y^2 - 7y^2 + 30y - 15y + 25$

Combining like terms, we get:

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

So, our quadratic equation is $2y^2 + 15y + 25 = 0$. Now that we have the equation in standard form, we can attempt to solve it. The next step typically involves trying to factor the quadratic. If factoring proves difficult, we can resort to the quadratic formula, which provides a guaranteed method for finding the solutions of any quadratic equation.

Step 4: Solving the Quadratic Equation

We have the quadratic equation $2y^2 + 15y + 25 = 0$. Now, we need to find the values of y that satisfy this equation. The first approach is often to try factoring the quadratic expression. Factoring involves expressing the quadratic as a product of two binomials. If factoring is not straightforward, we can use the quadratic formula, which is a general method for solving quadratic equations of the form $ay^2 + by + c = 0$. The quadratic formula is given by:

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

In our case, a = 2, b = 15, and c = 25. Let's first try to factor the quadratic expression. We are looking for two numbers that multiply to $2 * 25 = 50$ and add up to 15. These numbers are 10 and 5. So, we can rewrite the middle term as $10y + 5y$:

$2y^2 + 10y + 5y + 25 = 0$

Now, we factor by grouping:

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

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

Setting each factor equal to zero gives the solutions:

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

Solving for y:

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

Thus, we have found two potential solutions for y. The next critical step is to check these solutions in the original equation to ensure they are valid and do not introduce any extraneous roots. Extraneous roots can occur when we square both sides of an equation, as this operation can sometimes create solutions that do not satisfy the original equation.

Step 5: Checking for Extraneous Solutions

We have found two potential solutions for the equation $\sqrt{7y^2 + 15y} - 2y = 5 + y$: $y = -\frac{5}{2}$ and $y = -5$. It is crucial to verify these solutions by substituting them back into the original equation. This step is necessary because squaring both sides of an equation can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. We must ensure that our solutions make the original equation true and that the expressions under the square root remain non-negative.

First, let's check $y = -\frac{5}{2}$:

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

$\sqrt{7(\frac{25}{4}) - \frac{75}{2}} + 5 = \frac{5}{2}$

$\sqrt{\frac{175}{4} - \frac{150}{4}} + 5 = \frac{5}{2}$

$\sqrt{\frac{25}{4}} + 5 = \frac{5}{2}$

$\frac{5}{2} + 5 = \frac{5}{2}$

$\frac{15}{2} \neq \frac{5}{2}$

So, $y = -\frac{5}{2}$ is an extraneous solution.

Now, let's check $y = -5$:

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

$\sqrt{7(25) - 75} + 10 = 0$

$\sqrt{175 - 75} + 10 = 0$

$\sqrt{100} + 10 = 0$

$10 + 10 = 0$

$20 \neq 0$

So, $y = -5$ is also an extraneous solution.

Conclusion

After solving the equation $\sqrt{7y^2 + 15y} - 2y = 5 + y$ and checking the potential solutions, we found that both $y = -\frac{5}{2}$ and $y = -5$ are extraneous solutions. This means that neither of these values satisfies the original equation. Therefore, the equation has no real solutions. This highlights the importance of checking solutions in the original equation, especially when dealing with radical equations. Squaring both sides can introduce extraneous solutions, and it is essential to verify each solution to ensure it is valid.

Therefore, the correct answer is:

E. The equation has no solutions.