Problem 20 - Precalculus

Question

Solve the equation or inequality. $$3 x+\sqrt{6-9 x}=2$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{2}{3}\).

Step 1: Simplify the Equation

Let's start by isolating the term containing the square root. Given equation: \(3x + \sqrt{6 - 9x} = 2\). Subtract \(3x\) from both sides to get: \(\sqrt{6 - 9x} = 2 - 3x\).

Step 2: Eliminate the Square Root

Square both sides of the equation to eliminate the square root. This gives: \((\sqrt{6 - 9x})^2 = (2 - 3x)^2\). Simplifying, you get \(6 - 9x = 4 - 12x + 9x^2\).

Step 3: Simplify and Rearrange the Equation

Rearrange the equation by bringing all terms to one side: \(9x^2 - 3x - 2 = 0\). This is a quadratic equation in the form \(ax^2 + bx + c = 0\) with coefficients \(a=9\), \(b=-3\), \(c=-2\).

Step 4: Solve the Quadratic Equation Using the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 9\), \(b = -3\), and \(c = -2\) into the formula: \(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4\cdot9\cdot(-2)}}{2\cdot9}\).

Step 5: Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\): \((-3)^2 - 4(9)(-2) = 9 + 72 = 81\).

Step 6: Compute the Solution

Since the discriminant is positive, calculate the solutions: \(x = \frac{3 \pm \sqrt{81}}{18}\). This gives two potential solutions, \(x = \frac{3 + 9}{18} = \frac{12}{18} = \frac{2}{3}\) and \(x = \frac{3 - 9}{18} = \frac{-6}{18} = -\frac{1}{3}\).

Step 7: Check for Extraneous Solutions

Substitute each potential solution back into the original equation to verify. For \(x = \frac{2}{3}\): \(3(\frac{2}{3}) + \sqrt{6 - 9(\frac{2}{3})} = 2\) simplifies to \(2 + 0 = 2\), which is correct. For \(x = -\frac{1}{3}\): \(3(-\frac{1}{3}) + \sqrt{6 - 9(-\frac{1}{3})} = 2\) simplifies to \(-1 + \sqrt{9} = 2\), which is incorrect since \(2 eq 2\). Only \(x = \frac{2}{3}\) is a valid solution.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). Here, \( a, b, \) and \( c \) are constants, where \( a eq 0 \). The equation represents a parabola when graphed on a coordinate plane. Quadratics are fundamental in algebra and arise in various contexts including physics, engineering, and many other fields.

• The equation is always solved for \( x \), the variable representing the unknown value.
• The solution can be found using different methods: factoring, completing the square, or using the quadratic formula.
• The solutions are called the "roots" of the equation and can be real or complex numbers.

In our solution, we identified the quadratic equation as \( 9x^2 - 3x - 2 = 0 \), and we proceeded to solve it using the quadratic formula which leads us to find potential real solutions for this problem.

Square Roots

Square roots involve finding a number that, when multiplied by itself, gives the original number. The square root symbol is \( \sqrt{} \). Understanding square roots is crucial in algebra, especially when dealing with quadratic equations.

• To eliminate a square root, you square both sides of the equation.
• This process can sometimes introduce extraneous solutions, which necessitates verification of the potential solutions.
• Square roots are central to solving equations that involve complex expressions under the root symbol.

In the exercise, we eliminated the square root of \( 6 - 9x \) by squaring both sides of the equation, which transformed it into a form that made it easier to solve as a quadratic equation.

Extraneous Solutions

Extraneous solutions are results of solving an equation that don't satisfy the original equation. They typically arise when both sides of an equation are squared or when operations that expand the solution set are used.

• Checking solutions involves substituting them back into the original equation to ensure they satisfy it.
• This checking process is essential to verify the correctness because squaring an equation can introduce false solutions.
• In the given step-by-step solution, while checking substitute solutions, one solution did not satisfy the original equation, thus identified as extraneous.

Always remember to verify all solutions obtained, especially after manipulating equations involving square roots or powers.

Discriminant

The discriminant is a valuable component of the quadratic formula that can determine the nature and number of solutions for a quadratic equation. It is expressed as \( b^2 - 4ac \) in the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The discriminant tells us:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If the discriminant is negative, the solutions are complex and not real numbers.

In the example, the discriminant was calculated as 81, which is positive, thereby indicating two real solutions. However, only one of the solutions, upon verification, proved to be valid, demonstrating the introduction of extraneous solutions.