Problem 91 - Elementary and Intermediate Algebra

Question

Find the distance between each pair of points. \((\sqrt{48}, \sqrt{150})\) and \((\sqrt{12}, \sqrt{24})\)

Step-by-Step Solution

Answer

The distance between the points is \(\sqrt{66}\).

Step 1: Identify the Formula

To find the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\), use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Substitute the Values

Substitute the given coordinates \((\sqrt{48}, \sqrt{150})\) and \((\sqrt{12}, \sqrt{24})\) into the formula. Thus, \(x_1 = \sqrt{48}\), \(y_1 = \sqrt{150}\), \(x_2 = \sqrt{12}\), \(y_2 = \sqrt{24}\).Replace the variables in the formula to get:\[ d = \sqrt{(\sqrt{12} - \sqrt{48})^2 + (\sqrt{24} - \sqrt{150})^2} \]

Step 3: Simplify the Radicals

Simplify the square roots:- \( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \)- \( \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6} \)- \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \)- \( \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} \)

Step 4: Calculate Differences of the Coordinates

Compute the differences:- \( \sqrt{12} - \sqrt{48} = 2\sqrt{3} - 4\sqrt{3} = -2\sqrt{3} \)- \( \sqrt{24} - \sqrt{150} = 2\sqrt{6} - 5\sqrt{6} = -3\sqrt{6} \)

Step 5: Square the Differences

Square the differences computed earlier:- \((-2\sqrt{3})^2 = 4 \times 3 = 12\)- \((-3\sqrt{6})^2 = 9 \times 6 = 54\)

Step 6: Sum and Take the Square Root

Add the squared differences and take the square root to find the distance:\[d = \sqrt{12 + 54} = \sqrt{66}\]

Step 7: Simplify the Result

Although \(\sqrt{66}\) cannot be simplified further into an integer, it can be left as is or approximated as a decimal for further use as necessary.

Key Concepts

Simplifying Radicals

Simplifying radicals is a crucial step when working with square roots in mathematics. In this exercise, we encounter square roots like \(\sqrt{48}\) and \(\sqrt{150}\). Simplifying these helps in making the expressions easier to manage. To simplify a radical:

• Look for perfect square factors within the number under the square root. For instance, 48 can be broken down into 16 and 3, where 16 is a perfect square.
• Extract the square root of the perfect square factor. For \(\sqrt{48}\), this means \(\sqrt{16 \times 3} = 4\sqrt{3}\).
• Apply the same process to other radicals: \(\sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6}\).

By simplifying radicals, you make further calculations, such as addition or subtraction of similar terms, more straightforward and manageable.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using algebra to study geometric problems. It allows us to calculate distances between points, slopes of lines, and other geometric properties using coordinates.
In our given problem, we use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula finds the "straight line" distance between two sets of coordinates \((x_1, y_1)\) and \((x_2, y_2)\). Essentially, it builds upon the Pythagorean theorem:

• The horizontal distance between the points is \((x_2 - x_1)\).
• The vertical distance is \((y_2 - y_1)\).
• The distance between the points is the hypotenuse of the right triangle formed by these two legs.

Understanding coordinate geometry helps spatially understand algebraic equations and their graphical meanings.

Square Roots Calculation

Calculating square roots is a widely used mathematical operation that, on its surface, might seem daunting. However, recognizing perfect squares and breaking them down into simpler forms can ease this process dramatically.
To calculate a square root:

• Identify if the number is a perfect square. For non-perfect squares, find the nearest perfect square factors.
• If the number is not a perfect square, approximate the square root if an exact value isn't necessary or required (such as \(\sqrt{66}\), which approximates to 8.124).
• State when further simplification isn't possible or practical, and express your answer in radical form.

For complex numbers requiring square roots, breakdowns into simpler components mean reduced errors and a tightened mathematical understanding.