Problem 61 - College Algebra and Calculus: An Applied Approach

Question

Use a calculator to find the real solutions of the equation. (Round your answers to three decimal places.) \(1.8 x-6 \sqrt{x}-5.6=0\)

Step-by-Step Solution

Answer

The solutions are \(x\) = ... and \(x\) = ..., rounded to three decimal places.

Step 1: Identify a, b, and c

By comparing the equation \(1.8 x-6 \sqrt{x}-5.6=0\) with the general form of a quadratic equation \(ax^2+bx+c=0\), let \(x = u^2\), then the original equation becomes \(1.8 u^2-6u-5.6=0\). Here \(a = 1.8\), \(b = -6\), and \(c = -5.6\).

Step 2: Apply the quadratic formula

The quadratic formula is \[ u = \frac{-b±\sqrt{b^2-4ac}}{2a} \]. Substituting the values we obtained before, we get \(u = \frac{-(-6)±\sqrt{(-6)^2-4*1.8*(-5.6)}}{2*1.8}\).

Step 3: Solve the equation

Use the calculator to find the values for \(u\), remembering that \(u = \sqrt{x}\). Using the quadratic formula, we get two potential solutions for \(u\).

Step 4: Check the validity and square the result

Check if these solutions are indeed real and valid by substituting them back into the original equation. If the solutions are valid, square them to get the solutions for \(x\). Real solutions are valid only when \(\sqrt{x}\) is nonnegative.

Key Concepts

Understanding Quadratic Equations

Quadratic equations form the basis for many mathematical problems and scenarios. They take the general form: - \( ax^2 + bx + c = 0 \).This type of equation is characterized by the highest exponent of the variable being 2, thus the term "quadratic."
In the given exercise, the equation looks somewhat different because of the presence of a square root term. It's initially written as \( 1.8x - 6\sqrt{x} - 5.6 = 0 \).
To make it easier to solve, we substitute \( x = u^2 \), transforming the equation into a standard quadratic form: \( 1.8u^2 - 6u - 5.6 = 0 \). Now, it's clear how this relates to the standard quadratic equation format, with \( a = 1.8 \), \( b = -6 \), and \( c = -5.6 \).
These transformations are crucial in simplifying the equation so we can apply further techniques to find solutions.

Using Calculators to Solve Equations

Calculators are invaluable tools for solving complex equations like our quadratic example here. They help to perform computations that would be tedious and error-prone by hand, particularly when it involves square roots and fractions.
Once we've transformed our equation into a more manageable form, we can input the coefficients \( a \), \( b \) and \( c \) into a scientific calculator to find the roots. Calculators come equipped with specific functions for solving quadratics, often under a 'solve' or similar function.

• Input your coefficients: Enter \( a = 1.8 \), \( b = -6 \), and \( c = -5.6 \).
• Use the quadratic formula option if available.
• Note that results may need to be rounded. In this exercise, solutions should be rounded to three decimal places.

Always double-check your input values and ensure your calculator is set to the correct mode to avoid mistakes.

Applying the Quadratic Formula

The quadratic formula is a powerful method for finding solutions to any quadratic equation. It is given by:- \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this equation, \( b^2 - 4ac \) is called the discriminant. It helps determine the nature of the roots:

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

For our transformed equation \( 1.8u^2 - 6u - 5.6 = 0 \), substituting for \( a = 1.8 \), \( b = -6 \), and \( c = -5.6 \) provides real roots since the discriminant is positive.
Always remember, because we started with a \( \sqrt{x} \) term, calculating \( u \) gives us \( \sqrt{x} \), meaning you need to square \( u \) to find the final solutions for \( x \). This step ensures the solution is relevant to the original problem form.

Verifying Solutions

Verification is the final step in problem-solving, confirming that the solutions to our quadratic equation are indeed correct.
Once you've used the calculator to derive potential values for \( u \), each result must be checked by plugging back into the original equation \( 1.8x - 6\sqrt{x} - 5.6 = 0 \) to ensure it holds true.
This process helps to:

• Confirm the mathematical validity of our solutions.
• Identify any computational errors.
• Ensure that \( \sqrt{x} \) remains nonnegative, as the square root of a number should not be negative in real solutions.

If the substituted value validates the equation, then you're assured that your value for \( x \) is correct. This rigorous checking is crucial, especially in the case of rounding decisions or initial assumptions made during the problem transformation phase. Always review each step and consider all assumptions to ensure a thorough and accurate solution verification.