Problem 28 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \sqrt{4-6 x}=2 x $$

Step-by-Step Solution

Answer

There are no real solutions.

Step 1: Understand the Equation

We have the equation \( \sqrt{4-6x} = 2x \). This includes a square root on the left side and a linear term on the right side. Our goal is to find all real values of \(x\) that satisfy this equation.

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation: \((\sqrt{4-6x})^2 = (2x)^2\), which simplifies to \(4-6x = 4x^2\).

Step 3: Rearrange the Equation

Move all terms to one side to form a quadratic equation: \(4x^2 + 6x - 4 = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula where \(a = 4\), \(b = 6\), and \(c = -4\). The formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 5: Calculate the Discriminant

The discriminant \(b^2 - 4ac\) is calculated as \(6^2 - 4 \times 4 \times (-4) = 36 + 64 = 100\). Since the discriminant is positive, there are two real solutions.

Step 6: Compute the Solutions

Substitute the discriminant and other values into the quadratic formula: \(x = \frac{-6 \pm \sqrt{100}}{8}\). This gives \(x = \frac{-6 + 10}{8}\) and \(x = \frac{-6 - 10}{8}\).

Step 7: Simplify the Solutions

The solutions simplify to \(x = \frac{4}{8} = \frac{1}{2}\) and \(x = \frac{-16}{8} = -2\).

Step 8: Check for Extraneous Solutions

Plug \(x = \frac{1}{2}\) and \(x = -2\) back into the original equation to check. For \(x = \frac{1}{2}\), \(\sqrt{4 - 6(\frac{1}{2})} = \sqrt{1} = 1\) does not equal \(2(\frac{1}{2}) = 1\). For \(x = -2\), \(\sqrt{4 - 6(-2)} = \sqrt{16} = 4\) equals \(2(-2) = -4\), which is not valid since a square root cannot yield a negative number. Thus, there are no real solutions.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation where the highest degree of any term is 2. In general form, a quadratic equation can be written as

• \( ax^2 + bx + c = 0 \)

where \(a, b,\) and \(c\) are constants and \(a eq 0\). The solution to a quadratic equation can yield up to two real or complex roots. In the equation derived from the original problem,

• \(4x^2 + 6x - 4 = 0\),

we identify \(a = 4\), \(b = 6\), and \(c = -4\). This equation can be solved using various methods such as factorization, completing the square, or the quadratic formula.

Extraneous Solutions

Extraneous solutions are solutions that emerge from the process of solving equations but do not satisfy the original equation. These often occur when both sides of an equation are altered, such as when squaring both sides to eliminate a square root.

In the context of our problem, after squaring both sides of the equation \( \sqrt{4 - 6x} = 2x \) to form a quadratic equation, we derived solutions \( x = \frac{1}{2} \) and \( x = -2 \). However, checking these solutions against the original equation reveals that neither \( x = \frac{1}{2} \) nor \( x = -2 \) satisfy it, making them extraneous solutions.

Square Root

A square root is a value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because \(3 \times 3 = 9\). The symbol for a square root is \( \sqrt{} \).

In our problem, the expression \( \sqrt{4 - 6x} = 2x \) includes a square root. When dealing with square roots in equations, squaring both sides is a common technique used to simplify the expression, allowing us to isolate and solve for unknown variables. However, it is essential to verify potential solutions against the original equation because the squaring process can introduce extraneous solutions.

Discriminant

The discriminant is a component of the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) and determines the nature and number of the roots of a quadratic equation.

The discriminant is given as \( b^2 - 4ac \). In our equation, with \( a = 4 \), \( b = 6 \), and \( c = -4 \), the discriminant calculates to

• \( 6^2 - 4 \times 4 \times (-4) = 36 + 64 = 100 \).

Since the discriminant is positive, it indicates that the quadratic equation has two distinct real roots. This information is crucial in predicting the number of solutions before actually solving the quadratic equation.