Problem 39 - College Algebra and Calculus: An Applied Approach

Question

Solve the quadratic equation by extracting square roots. List both the exact answer and a decimal answer that has been rounded to two decimal places. $$ 6 x^{2}-3\left(x^{2}+1\right)=23 $$

Step-by-Step Solution

Answer

The roots of the quadratic equation are \(x=\pm \frac{\sqrt{78}}{3}\) or in decimal form, \(x=\pm 5.20\).

Step 1: Simplify the equation

Firstly, simplify the equation by spreading out the multiplication on the right side of the equation. In that case, this equation simplifies to \(6x^{2}-3x^{2}-3=23\). Combine like terms to get \(3x^{2}-3=23\).

Step 2: Isolate the variable

Next we want to isolate the variable. To do this, first add 3 to both sides of the equation. Afterwards, the equation becomes \(3x^{2}=26\). Divide both sides by 3 to solve for \(x^{2}\), and we get \(x^{2}=\frac{26}{3}\).

Step 3: Solve for X

Now we want to solve for x. Since \(x^{2}=\frac{26}{3}\), take the square root on both sides will give us \(x=\pm \sqrt \frac{26}{3}\), extracting the root of 3 to simplify, we get \(x=\pm \frac{\sqrt{78}}{3}\).

Step 4: Convert to decimal form

Finally, convert the obtained result into decimal format. Thus, using a calculator to round off to 2 decimal places we get \(x=\pm 5.20\).

Key Concepts

Extracting Square Roots

Extracting square roots is one of the simplest methods for solving quadratic equations where the highest power of the variable is squared. In this context, it is useful when the equation can be rewritten in the basic form of \(x^2 = a\), where 'a' is a number. The square root of both sides is then taken to find values for the variable 'x'.

In our example, after simplifying and isolating the variable, the equation \(x^2 = \frac{26}{3}\) is ready for the square root to be extracted. By applying the square root to both sides, we arrive at two solutions: \(x = +\sqrt{\frac{26}{3}}\) and \(x = -\sqrt{\frac{26}{3}}\), because squaring either a positive or negative number results in a positive value, and thus we must consider both roots. This approach provides us with the exact solutions for 'x'.

Quadratic Formula

The quadratic formula is a versatile method for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It states that the solutions are given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where 'a', 'b', and 'c' are coefficients from the equation.

Even though the current problem does not require the use of the quadratic formula (as extracting square roots suffices), it's important to know that this formula can solve equations that are not easily factored or when extracting square roots isn't straightforward. In case our equation did not simplify nicely, the quadratic formula would be our tool to ensure we get the correct solutions.

Algebraic Simplification

Algebraic simplification is the process of combining like terms and simplifying expressions as much as possible. Often in quadratic equations, simplification is a crucial first step to make the equation solvable by methods like extracting square roots or applying the quadratic formula.

In the exercise provided, simplification involved expanding the multiplication and then combining like terms. Simplifying the equation from \(6x^2 - 3(x^2 + 1) = 23\) to \(3x^2 - 3 = 23\) makes it manageable. This process can sometimes reveal a simpler method to solve the equation, and in doing so, reduces the chance of making errors in subsequent steps.

Decimal Approximation

Decimal approximation is used to express algebraic solutions in a more practical, decimal form, often required when exact forms are either too complex or not necessary for the application. Real-world problems frequently call for numerical approximations, hence knowing how to convert an exact algebraic solution into a decimal answer is valuable.

For the given exercise, the exact answers are \(x = \pm \frac{\sqrt{78}}{3}\). However, for practical purposes, these answers are approximated to two decimal places using a calculator, yielding \(x = \pm 5.20\). The act of rounding to a certain number of decimal places should always be performed with consideration of the desired precision and the context in which the answer will be used.