Problem 58 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the equation. $$ \sqrt{6 x^{2}-5 x}-2=0 $$

Step-by-Step Solution

Answer

The solutions for the equation \(\sqrt{6x^2-5x}-2=0\) are \(x_1 = \frac{4}{3}\) and \(x_2 = -\frac{1}{2}\).

Step 1: Isolate the square root

To isolate the square root, we need to move the constant term, 2, to the other side of the equation: $$ \sqrt{6x^2-5x} = 2 $$

Step 2: Square both sides of the equation

Squaring both sides of the equation will eliminate the square root. Be careful to square both sides, not just the square root: $$ (\sqrt{6x^2-5x})^2 = (2)^2 $$

Step 3: Simplify the equation

Now, simplify the equation: $$ 6x^2-5x=4 $$

Step 4: Bring all terms to one side of the equation

Move all terms to one side to create a quadratic equation: $$ 6x^2-5x-4 = 0 $$

Step 5: Solve the quadratic equation

To solve the quadratic equation, use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Where a = 6, b = -5, and c = -4: $$ x = \frac{5 \pm \sqrt{(-5)^2 - 4(6)(-4)}}{2(6)} $$

Step 6: Calculate the discriminant

Calculate the discriminant (the value inside the square root): $$ D = (-5)^2 - 4(6)(-4) = 25 + 96 = 121 $$

Step 7: Calculate the values of x

Now, insert the discriminant value into the quadratic formula and calculate the values of x: $$ x = \frac{5 \pm \sqrt{121}}{12} = \frac{5 \pm 11}{12} $$ There are two possible solutions for x: $$ x_1 = \frac{5 + 11}{12} = \frac{16}{12} = \frac{4}{3} \\ \\ x_2 = \frac{5 - 11}{12} = \frac{-6}{12} = -\frac{1}{2} $$ So, the solutions for the equation \(\sqrt{6x^2-5x}-2=0\) are \(x_1 = \frac{4}{3}\) and \(x_2 = -\frac{1}{2}\).

Key Concepts

Square Root Equations

Understanding square root equations is key to solving problems like the one presented in the exercise. Square root equations typically involve a square root expression set equal to a number, as seen in our given equation: \(\sqrt{6x^2 - 5x} - 2 = 0\).
To tackle these, your first aim is to isolate the square root. This means getting the radical by itself on one side of the equation. Here, we moved the \(-2\) to the other side, leaving us with: \[ \sqrt{6x^2 - 5x} = 2\].

• Step-by-step progressing to isolate the square root is crucial to simplify the expression.
• Once isolated, squaring both sides of the equation will eliminate the square root.
• This method helps transform the equation into a more familiar form, often quadratic.

Squaring solves the radical, but remember to check potential solutions in the original equation. Some may not actually satisfy the original because squaring can introduce extraneous solutions.

Discriminant

The discriminant is a fundamental part of the quadratic formula, helping us determine the nature and number of solutions in a quadratic equation. The discriminant is calculated from the formula:\( b^2 - 4ac \).
In the equation obtained in our exercise, 6 is the coefficient \(a\), -5 is \(b\), and -4 is \(c\). Substituting these values:\(D = (-5)^2 - 4(6)(-4)\).
This results in \(D = 25 + 96 = 121\).

• If \(D > 0\), there are two real and distinct solutions, meaning the quadratic will intersect the x-axis at two separate points.
• If \(D = 0\), there's exactly one real solution, where the graph barely touches the x-axis.
• If \(D < 0\), the solutions are complex or non-real, indicating the curve does not intersect the x-axis.

In our problem, because \(D = 121\) (a perfect square), we discovered there are two real solutions, fostering confidence in our calculations before applying the quadratic formula.

Quadratic Formula

The quadratic formula offers a powerful method to find the solutions of a quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
This formula is derived from completing the square on a quadratic equation and provides solutions in terms of the coefficients \(a\), \(b\), and \(c\).

• The \(+/-\) in the formula accounts for the two potential solutions in cases where the discriminant is positive.
• After inserting the correct values into the formula, solve the expression to find the roots of the equation.
• The roots we calculated were \( x_1 = \frac{4}{3} \) and \( x_2 = -\frac{1}{2} \), aligning perfectly with known mathematical principles.

These solutions needed to satisfy the conditions that our original manipulated equation, \(6x^2 - 5x = 4\), mandates. Thus, the quadratic formula is versatile, providing both efficiency and accuracy in solving quadratic equations.