Problem 28 - Algebra 2 and Trigonometry

Question

In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ 3+\sqrt{4 x-3}=2 x $$

Step-by-Step Solution

Answer

The solution set is \( \{3\} \).

Step 1: Isolate the Square Root

Start by isolating the square root term on one side of the equation. The original equation is: \[ 3 + \sqrt{4x-3} = 2x \]Subtract 3 from both sides to get:\[ \sqrt{4x-3} = 2x - 3 \]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation:\[ (\sqrt{4x-3})^2 = (2x - 3)^2 \]This simplifies to:\[ 4x - 3 = 4x^2 - 12x + 9 \]

Step 3: Rearrange and Simplify

Bring all terms to one side of the equation to form a quadratic equation:\[ 4x^2 - 12x + 9 - 4x + 3 = 0 \]Simplify it:\[ 4x^2 - 16x + 12 = 0 \]

Step 4: Factor the Quadratic Equation

Factor the quadratic equation:2x can be taken common:\[ 4(x^2 - 4x + 3) = 0 \] This factors further into:\[ 4(x - 3)(x - 1) = 0 \]

Step 5: Solve for x

Set each factor equal to zero and solve for \(x\):1. \(x - 3 = 0\) gives \(x = 3\)2. \(x - 1 = 0\) gives \(x = 1\)

Step 6: Check Each Solution

Substitute \(x = 3\) back into the original equation:\[ 3 + \sqrt{4(3) - 3} = 2(3) \]\[ 3 + \sqrt{9} = 6 \]\[ 3 + 3 = 6 \]It checks out, so \(x = 3\) is a solution.Now check \(x = 1\):\[ 3 + \sqrt{4(1) - 3} = 2(1) \]\[ 3 + \sqrt{1} = 2 \]\[ 3 + 1 = 2 \]This is incorrect, so \(x = 1\) is not a solution.

Step 7: Write the Solution Set

The only solution that satisfies the original equation is \(x = 3\). Therefore, the solution set is:\[ \{3\} \]

Key Concepts

Factorization

Factorization is a key step when solving quadratic equations. It transforms a complex polynomial into a product of simpler binomials. This makes finding solutions much easier. Consider the equation obtained after removing the square root from: \[ 4x^2 - 16x + 12 = 0 \] This quadratic equation can be factored by taking out the greatest common factor which in this case is 4: \[ 4(x^2 - 4x + 3) = 0 \]

• Next, focus on the trinomial \(x^2 - 4x + 3\). Look for two numbers that multiply to 3 and add to -4.
• These numbers are -3 and -1. Hence, the expression can be factored as \((x - 3)(x - 1)\).

Thus, the factorization turns the quadratic into simple linear expressions: \(4(x - 3)(x - 1) = 0\). Factoring allows for a straightforward path to find each possible solution.

Solution Sets

A solution set is a collection of values that satisfy an equation. After factorizing the quadratic equation into \((x - 3)(x - 1) = 0\), each bracket can be solved separately:

• Setting \(x - 3 = 0\), we find \(x = 3\).
• Likewise, setting \(x - 1 = 0\), gives \(x = 1\).

In theory, both values are potential solutions. However, not every potential solution is correct. Therefore, it’s crucial to verify each separately by substituting back into the original equation. This leads us to the concept of checking solutions.

Checking Solutions

Checking solutions is an essential step in solving equations, as it determines which potential solutions are valid. Consider our findings of \(x = 3\) and \(x = 1\):

• To verify, substitute \(x = 3\) into the original equation \(3 + \sqrt{4x-3} = 2x\). It simplifies to \(6 = 6\), confirming \(x = 3\) is a solution.
• Substitute \(x = 1\), resulting in \(4 eq 2\). Thus, \(x = 1\) is not a solution.

Checking helps ensure that the solution set only contains values that genuinely satisfy the original quadratic equation.

Quadratic Expression

A quadratic expression is a polynomial of degree 2. It typically takes the form \(ax^2 + bx + c\). In our problem, we dealt with the quadratic \(4x^2 - 16x + 12\). Understanding the components of quadratic expressions is crucial:

• The term \(ax^2\) represents the quadratic or squared term, influencing the shape of the parabola.
• The term \(bx\) is the linear term, affecting the direction the parabola opens.
• The constant term \(c\) shifts the graph vertically.

By examining these components, we form the basis for factorization and solving quadratic equations effectively. Quadratic expressions are fundamental in algebra and appear in various real-world scenarios such as physics and engineering.