Problem 29 - College Algebra and Calculus: An Applied Approach

Question

Find the real solution(s) of the radical equation. Check your solutions. \(x=\sqrt{11 x-30}\)

Step-by-Step Solution

Answer

The solutions to the radical equation \(x = \sqrt{11x - 30}\) are \(x = 5\) and \(x = 6\).

Step 1: Square both sides of the equation

In order to remove the radical (square root), square both sides of the equation. This gives the equation \(x^2 = 11x - 30\).

Step 2: Rearrange to form a quadratic equation

Subtract \(11x\) from both sides to get all terms on one side of the equation. Add 30 to both sides as well to achieve a standard quadratic equation format. This gives us the equation \(x^2 - 11x + 30 = 0\).

Step 3: Factor the quadratic equation

Factoring the quadratic equation gives \((x - 5)(x - 6) = 0\).

Step 4: Solve for x

Setting each factor of the equation equal to zero gives us \(x - 5 = 0\) or \(x - 6 = 0\), which leads to two possible solutions for \(x\), i.e., \(x = 5\) and \(x = 6\).

Step 5: Check the solutions

Substitute these values back into the original equation to check if they are valid solutions. Upon substituting, it is found that both \(x = 5\) and \(x = 6\) are valid solutions to the equation.

Key Concepts

Understanding Quadratic Equations

In the context of solving radical equations, the concept of a quadratic equation becomes central once we have eliminated the square root. A quadratic equation is a second-degree polynomial, which generally takes the form:\[ax^2 + bx + c = 0\]Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). The power of 2 in \(x^2\) signifies that it is quadratic, as opposed to linear (which would just involve \(x\)).Quadratic equations are familiar due to their characteristic `U`-shaped graph called a parabola. It's crucial to identify when an equation can be rewritten as a quadratic form. In our problem, by eliminating the square root, the expression \(x = \sqrt{11x - 30}\) is transformed into the quadratic equation \(x^2 = 11x - 30\). Rearranging further gives us \(x^2 - 11x + 30 = 0\), highlighting the utility of various algebraic skills in such transformations.

The Power of Factoring

Factoring is a potent technique used to solve quadratic equations. It involves rewriting the equation as a product of its factors, which can make finding solutions much simpler.To factor a quadratic like \(x^2 - 11x + 30 = 0\), we're looking for two numbers that multiply to the constant term \(30\) and add up to the linear coefficient, \(-11\). In this case, \(-5\) and \(-6\) satisfy both conditions:

• The product \((-5) \times (-6) = 30\)
• The sum \(-5 + (-6) = -11\)

Thus, the equation factors into:\[(x - 5)(x - 6) = 0\]Once the equation is broken into factors, we apply the "zero product property". This property states that if the product of two terms is zero, then at least one of the terms must be zero. Therefore, we set each factor equal to zero, solving for \(x\):

• \(x - 5 = 0\), leading to \(x = 5\)
• \(x - 6 = 0\), leading to \(x = 6\)

Solution Verification

Solution verification is a critical step, especially when solving equations involving radicals. Checking potential solutions ensures they satisfy the original equation.For our original equation \(x = \sqrt{11x - 30}\), we found the possible solutions \(x = 5\) and \(x = 6\). Verification involves substituting these values back into the equation to confirm they hold true:

• Substitute \(x = 5\): - Left side: \(5\) - Right side: \(\sqrt{11 \times 5 - 30} = \sqrt{25} = 5\) - Both sides are equal, proving \(x = 5\) is valid.
• Substitute \(x = 6\): - Left side: \(6\) - Right side: \(\sqrt{11 \times 6 - 30} = \sqrt{36} = 6\) - Both sides are equal, proving \(x = 6\) is valid.

Verification is particularly important for radical equations since manipulating them (e.g., squaring both sides) can introduce extraneous solutions. Therefore, always check potential solutions against the initial equation to confirm validity.