Question

Solve for \(x\) and check. $$\frac{6}{\sqrt{3+x}}=\sqrt{x+3}$$

Short answer

The solution for \(x\) is 3 after checking for extraneous solutions.

Step-by-step solution

Square both sides

To remove the square root, square both sides of the equation: \(\left(\frac{6}{\sqrt{3+x}}\right)^2 = (\sqrt{x+3})^2\).

Simplify both sides

Simplify the equation by squaring the terms: \(\frac{36}{3+x} = x+3\).

Multiply both sides by \(3+x\)

To clear the denominator, multiply both sides of the equation by \(3+x\): \(36 = (x+3)(3+x)\).

Expand the right-hand side

Expand the right side of the equation: \(36 = x^2 + 3x + 3x + 9\).

Simplify the right-hand side

Combine like terms on the right side: \(36 = x^2 + 6x + 9\).

Set the equation to zero

Subtract 36 from both sides to set the quadratic equation to zero: \(x^2 + 6x + 9 - 36 = 0\).

Simplify the equation

Combine like terms to simplify the equation: \(x^2 + 6x - 27 = 0\).

Factor the quadratic equation

Factor the quadratic equation to find the values of \(x\): \(x + 9)(x - 3) = 0\).

Solve for \(x\)

Set each factor equal to zero and solve for \(x\): \(x + 9 = 0\) or \(x - 3 = 0\), giving the solutions \(x = -9\) and \(x = 3\).

Check the solutions

Substitute the values of \(x\) back into the original equation to check for extraneous solutions. Only \(x = 3\) is valid because it does not result in a negative number under the square root.

Key concepts

Quadratic Equations

A quadratic equation is a second-degree polynomial equation, generally of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \) and \( c \) are constants, with \( a \eq 0 \). Solving quadratic equations is a fundamental skill in algebra. There are multiple ways to find the roots, or solutions, of a quadratic equation, such as factorization, completing the square, and using the quadratic formula. In the given exercise, we encounter a quadratic equation after manipulating the original radical equation through various steps that involve squaring both sides and simplifying.

The process of squaring an equation can lead to a quadratic equation, which can then be simplified and set to zero, allowing us to use factorization to find the solutions. Factorization involves expressing the quadratic equation as a product of two binomials. If the quadratic is factorable, we can set each binomial equal to zero and solve for \( x \) to find the roots of the equation.

Factoring Polynomials

Factoring polynomials is a critical process in algebra that involves breaking down a polynomial into a product of smaller polynomials. For quadratic equations, which are a special case of polynomials, the standard form is \( ax^2 + bx + c \). Factoring is often used to find the roots of the equation, particularly when the quadratic can be easily decomposed into a product of binomial factors.

In our exercise, we arrived at a quadratic equation through carefully manipulated steps that transformed a radical equation into a form where factoring becomes relevant. To factor the quadratic obtained in the exercise, \( x^2 + 6x - 27 = 0 \), we look for two numbers that multiply to give the constant term (-27) while adding up to the coefficient of the \( x \) term, which is 6. This leads us to the factors \( x + 9 \) and \( x - 3 \) as the product must equal zero for the property of zero product to apply.

Extraneous Solutions

Extraneous solutions are results that emerge from the process of solving an equation but are not valid solutions to the original problem. These can arise when applying certain operations, such as squaring both sides of an equation, which can introduce additional roots. It's crucial to check all potential solutions by substiting them back into the original equation to confirm their validity.

The exercise we examined illustrates this concept perfectly. After solving the quadratic equation, we found two potential solutions for \( x \): -9 and 3. However, only \( x = 3 \) is a valid solution. When we substitute \( x = -9 \) back into the original radical equation, it becomes invalid because it generates a negative number under the square root, which is undefined in the realm of real numbers. This demonstrates how critical it is to always check your solutions against the original equation to ensure they are indeed correct.