Problem 80 - Intermediate Algebra

Question

Solve each equation. $$ (5 x+6)^{\frac{1}{2}}=x $$

Step-by-Step Solution

Answer

The solution is \(x = 6\).

Step 1: Eliminate the Square Root

Square both sides of the equation to eliminate the square root. This gives us:\[(5x + 6) = x^2\]

Step 2: Rearrange Terms

Rearrange the terms to set the equation to zero on one side:\[x^2 - 5x - 6 = 0\]

Step 3: Factor the Quadratic

Factor the quadratic equation. We look for two numbers that multiply to -6 and add to -5. The numbers -6 and 1 work:\[(x - 6)(x + 1) = 0\]

Step 4: Solve for x

Set each factor equal to zero and solve for \(x\):\[x - 6 = 0 \quad \text{or} \quad x + 1 = 0\]This gives solutions:\\(x = 6\) and \(x = -1\).

Step 5: Check Solutions in Original Equation

Substitute each solution back into the original equation to verify:- For \(x = 6\): \[(5(6) + 6)^{1/2} = 6 \rightarrow (36)^{1/2} = 6 \rightarrow 6 = 6 \] (valid solution)- For \(x = -1\): \[(5(-1) + 6)^{1/2} = -1 \rightarrow (1)^{1/2} = -1 \rightarrow 1 = -1 \] (invalid solution)Thus, \(x = 6\) is the only valid solution.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that are second-degree. This means the highest power of the variable in the equation is squared. They generally have the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). To solve quadratic equations, we can utilize various methods such as factoring, using the quadratic formula, completing the square, or graphically.Quadratic equations are significant because they can model a wide variety of real-world phenomena. Whether you are calculating the trajectory of a projectile, designing engineering structures, or analyzing financial investments, quadratic equations are invaluable. Solving them allows us to find these vital points where these calculations hold true.When you encounter a quadratic equation in any form, your first goal is usually to rearrange and simplify it so that it fits the form \( ax^2 + bx + c = 0 \). This set-up simplifies the equation and sets you up for the next step, which is often factorization.

Factorization

Factorization is a method used to solve quadratic equations by expressing the quadratic in a product form, where two binomials are multiplied to yield the original equation. The principle behind factorization is finding two numbers that multiply to the constant term and add up to the linear coefficient. Consider the quadratic equation from the previous example \( x^2 - 5x - 6 = 0 \).To factor this:

• We look for two numbers that multiply to \(-6\) (the constant term) and add up to \(-5\) (the coefficient of \(x\)).
• The numbers \(-6\) and \(1\) meet these conditions.

Upon identification, we can rewrite the equation as \( (x - 6)(x + 1) = 0 \). This factorization allows us to solve each part separately. Setting each factor equal to zero gives the potential solutions \( x = 6 \) and \( x = -1 \).Factorization not only helps in solving quadratic equations but also aids in understanding the roots or solutions of the equation. Each root corresponds to a point where the quadratic graph intersects or touches the x-axis.

Verification of Solutions

Verification is crucial in solving equations, as it ensures that the proposed solutions are indeed correct. Not all solutions derived from an equation might satisfy the original equation, especially in cases involving square roots, as extraneous solutions could appear. In our example equation, solved as \((5x + 6)^{\frac{1}{2}} = x\), we found \(x = 6\) and \(x = -1\) as potential solutions.Verifying involves substituting these values back into the original equation to see if they hold true. For \(x = 6\):\[(5(6) + 6)^{1/2} = 6 \rightarrow (36)^{1/2} = 6 \rightarrow 6 = 6\]This is a valid solution. However, for \(x = -1\):\[(5(-1) + 6)^{1/2} = -1 \rightarrow (1)^{1/2} = -1 \rightarrow 1 = -1\]This is not valid, as it contradicts the original equation.Verifying solutions safeguards against errors and ensures understanding of the solution's relevance to the initial equation. Always check your solutions, especially when dealing with powers and roots, to confirm their validity.