Problem 16 - Algebra 2

Question

Solve. Check for extraneous solutions. \((5 x+4)^{\frac{1}{2}}-3 x=0\)

Step-by-Step Solution

Answer

The solution to the equation are the those values of \(x\) which satisfy the original equation after being checked for extranity. If either or both of the solutions do not satisfy the original equation, they are considered extraneous and discarded.

Step 1: Isolate the Square Root Term

First, let's isolate the square root term on one side of the equation. To do this, we can move the term containing \(x\) to the right side by adding \(3x\) to both sides of the equation. This simplifies the equation to: \((5x + 4)^{\frac{1}{2}} = 3x\).

Step 2: Square Both Sides

To eliminate the radicals, we square both sides of the equation. Remember when we square a square root, we get the term under the square root. So we have: \((5x + 4) = (3x)^2\), which simplifies to: \(5x + 4 = 9x^2\).

Step 3: Rearrange the Equation to a Quadratic Form

We rearrange our equation into standard quadratic form, i.e., \(ax^2 + bx + c = 0\). This leads to: \(9x^2 - 5x + 4 = 0\).

Step 4: Solve the Quadratic Equation

We can solve this quadratic equation either by factoring, completing the square or the quadratic formula. However, this equation doesn't factorize easily, so let's use the quadratic formula: \(x = [-b ± \sqrt{b^2 - 4ac}]/(2a)\). After substituting a=9, b=-5, and c=4, we find the solutions for \(x\).

Step 5: Check for Extraneous Solutions

Substitute the calculated values of \(x\) into the original equation and verify whether they satisfy the equation. Discard any false solutions.

Key Concepts

Quadratic Equations

Quadratic equations are one of the most fundamental concepts in algebra. They typically have the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. In our exercise, the equation is transformed into the quadratic expression \(9x^2 - 5x + 4 = 0\). Here,

• \(a = 9\)
• \(b = -5\)
• \(c = 4\)

These parameters define the properties of the quadratic equation,
such as its shape and the nature of its roots. Quadratic equations can be solved using several methods, including:

• Factoring, if it can be factored easily.
• Completing the square, which rearranges the equation to find the roots.
• The quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

This formula provides the solutions to any quadratic equation,
even when factoring and completing the square don't apply.
In our solution, we used the quadratic formula because it allowed us to straightforwardly find the values of \(x\) that solve the quadratic equation.

Square Roots

The square root symbol \(\sqrt{}\) represents a value that, when multiplied by itself, gives the original number. In our problem, we begin with the term \((5x + 4)^{\frac{1}{2}}\),
which can also be rewritten as \(\sqrt{5x + 4}\). Solving equations with square roots often involves this crucial step:

• Isolating the square root expression on one side of the equation. In this exercise, it resulted in \((5x + 4)^{\frac{1}{2}} = 3x\).
• Squaring both sides to remove the square root. This yields \(5x + 4 = 9x^2\), eliminating the radical sign and simplifying the problem into a quadratic equation.

Understanding how to manipulate square roots is essential,
as they frequently appear in algebraic problems. By squaring both sides judiciously,
we effectively simplify the equation and prepare it for further analysis using methods like solving quadratic equations.

Extraneous Solutions

When solving radical equations, particularly those involving square roots,
extraneous solutions can sometimes arise.
These are solutions that emerge from the mathematical manipulation of the equation
but do not satisfy the original equation.
This discrepancy often occurs because squaring is a reversible process,
meaning squaring both sides of an equation might introduce solutions that don't actually work in the original.

• It's essential to substitute any solutions back into the original equation and verify their validity.
• If a solution doesn't hold true in the original statement, it's deemed extraneous and should be discarded.

In our problem, after finding the potential values for \(x\) using the quadratic formula,
each must be checked to ensure it does not introduce a falsehood unrelated to the logical structure of the original radical equation.
This validation step ensures that only true solutions remain, providing an accurate answer to the problem.