Problem 81 - Intermediate Algebra

Question

Solve each equation. $$ (3 x+4)^{\frac{1}{2}}=x $$

Step-by-Step Solution

Answer

The solution is \(x = 4\).

Step 1: Square Both Sides

To eliminate the square root, square both sides of the equation \[(3x + 4)^{\frac{1}{2}} = x\]. This yields:\[3x + 4 = x^2\].

Step 2: Rearrange the Equation

Bring all terms to one side of the equation to set it to zero:\[x^2 - 3x - 4 = 0\].

Step 3: Factor the Quadratic Equation

Find factors of the quadratic equation \[x^2 - 3x - 4\] that add up to \[-3\] and multiply to \[-4\].These factors are \[(x - 4)(x + 1) = 0\].

Step 4: Solve for x

Apply the zero-product property, which states that if a product of two factors is zero, then at least one of the factors must be zero. Therefore, solve the equations: \[x - 4 = 0\] and \[x + 1 = 0\].This gives the solutions \[x = 4\] and \[x = -1\].

Step 5: Check the Solutions

Substitute the solutions back into the original equation \[(3x + 4)^{\frac{1}{2}} = x\] to verify their validity:- For \(x = 4\), the left side is \(\sqrt{3(4) + 4} = \sqrt{16} = 4\), so it satisfies the equation.- For \(x = -1\), the left side is \(\sqrt{3(-1) + 4} = \sqrt{1} = 1\), which does not equal \(-1\).Thus, \(x = 4\) is the only valid solution.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation in which the highest power of the variable is 2. They typically take the form \[ ax^2 + bx + c = 0 \], where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). Quadratics can appear when you square both sides of an equation to eliminate radicals, much like in the original exercise here.

When simplifying the radical equation \[ (3x + 4)^{\frac{1}{2}} = x \], squaring each side resulted in the quadratic equation \[ x^2 - 3x - 4 = 0 \].

Quadratic equations can be solved using several methods:

• Factoring, if the equation is factorable.
• Completing the square.
• Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

The method chosen often depends on the specific form of the equation. In this particular case, factoring was used.

Zero-Product Property

The zero-product property is a vital tool when solving quadratic equations. This property asserts that if the product of two numbers is zero, then at least one of the numbers must be zero. In mathematical terms, if \(a \cdot b = 0\), then either \(a = 0\) or \(b = 0\) (or both).

After rearranging and factoring the quadratic equation \[x^2 - 3x - 4 = 0\] as \[(x - 4)(x + 1) = 0\], you apply the zero-product property to find that:

• \(x - 4 = 0\)
• \(x + 1 = 0\)

This leads to potential solutions: \(x = 4\) or \(x = -1\).

The beauty of the zero-product property is its simplicity and power in breaking down complex equations into solvable pieces.

Checking Solutions

Once potential solutions are found, checking each solution is crucial. This step ensures that solutions satisfy the original equation, especially if the process involved squaring both sides. Squaring can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original.

In our exercise:

• For \(x = 4\): Substituting back into the original equation gives \(\sqrt{3(4) + 4} = \sqrt{16} = 4\), confirming it's valid.
• For \(x = -1\): Substituting gives \(\sqrt{3(-1) + 4} = \sqrt{1} = 1\), which does not equal \(-1\). Hence, \(x = -1\) is invalid for the original equation.

Thus, the only valid solution is \(x = 4\).

Checking not only verifies correctness but also deepens understanding by ensuring the solution genuinely fits the problem.