Problem 34 - 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. $$ 6 x^{2}=250 $$

Step-by-Step Solution

Answer

The exact solutions are \(x = \sqrt{\frac{250}{6}}\) and \(x = -\sqrt{\frac{250}{6}}\). After rounding to two decimal places, the solutions are approximately \(x = 8.16\) and \(x = -8.16\).

Step 1: Isolate \(x^{2}\)

To begin with, let's isolate \(x^{2}\) by dividing both sides of the equation by 6. This will give us the formula \(x^{2} = \frac{250}{6}\).

Step 2: Take the Square Root of Both Sides

Taking the square root of both sides gives us the two possible solutions for x. \(x = \sqrt{\frac{250}{6}}\) and \(x = -\sqrt{\frac{250}{6}}\).

Step 3: Simplify the Square Root

Simplify \(x = \sqrt{\frac{250}{6}}\) and \(x = -\sqrt{\frac{250}{6}}\) by dividing the numerator (250) by the denominator (6). Round the results to two decimal places.

Step 4: Calculate the Values of \(x\)

After simplifying, use a calculator to find the values of \(x = \sqrt{\frac{250}{6}}\) and \(x = -\sqrt{\frac{250}{6}}\) to two decimal places.

Key Concepts

Extracting Square Roots

When you encounter a quadratic equation like \(6 x^{2}=250\), the method of extracting square roots is often the simplest way to find the solution. This approach works best when the quadratic equation contains a perfect square term and no linear term—in other words, when it's in the form of \(ax^{2} = c\).

To isolate \(x^2\), we start by dividing both sides by the coefficient of the \(x^2\) term—in this case, 6. This gives us \(x^2 = \frac{250}{6}\). Once isolated, we can deal with the square root. Remember, a square root has both a positive and a negative solution because both \(\sqrt{x}\times\sqrt{x}\) and \(-\sqrt{x}\times-\sqrt{x}\) result in the original value. Hence, we express the solution as \(x = \pm\sqrt{\frac{250}{6}}\).

Simplifying requires rationalizing the fraction, if possible, and then extracting the square root for both the positive and negative cases. This part is crucial—missing out on the negative solution is a common mistake, which can lead to a half-correct answer.

Quadratic Formula

At times when extracting square roots isn't applicable, such as when you have a full-blown quadratic equation with a non-zero linear term, or the numbers aren't nicely workable, that's when the quadratic formula comes to the rescue. It is stated as \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a\), \(b\), and \(c\) are the coefficients of the terms \(ax^2\), \(bx\), and \(c\) respectively in the quadratic equation \(ax^2+bx+c=0\).

Using it requires substituting the coefficients of your quadratic equation into the formula. The term under the square root sign, \(b^2-4ac\), is known as the discriminant. It determines the nature of the roots—whether they are real and distinct, real and equal, or complex. If the discriminant is zero, the roots are real and equal. If it’s positive, you'll have two real and distinct roots. However, if it's negative, the roots will be complex numbers.

Rounding Decimals

After finding the exact solutions to our equations, we often need to round decimals for a more practical, approximate answer. This is important when exact solutions are unwieldy or impossible to express as a simple decimal. Rounding to two decimal places involves looking at the third decimal value. If this number is five or higher, you round up the second decimal place. If it's four or lower, you leave the second decimal place as it is.

For instance, if our exact solution was \(x = \pm\sqrt{\frac{250}{6}}\), and the decimal form was approximately \(x = \pm6.708203932\) when calculated, we'd round this to \(x = \pm6.71\) up to two decimal places. Rounding is a practical mathematical skill across various sciences and financial calculations, as it helps simplify complex numbers into more digestible figures while maintaining a reasonable level of accuracy.