Problem 40
Question
The width of a rectangle is \(x\) and the length is \(\sqrt{x-1}\) . If the width is twice the length, what are the dimensions of the rectangle?
Step-by-Step Solution
Verified Answer
The dimensions of the rectangle are width = 2 and length = 1.
1Step 1: Define the Problem
We are given that the width of the rectangle is \(x\), and the length is \(\sqrt{x-1}\). We need to find the dimensions of the rectangle given that the width is twice the length.
2Step 2: Set Up the Equation
Since the width is twice the length, we set up the equation: \(x = 2 \times \sqrt{x-1}\). This equation relates the width and length based on the given condition.
3Step 3: Isolate the Square Root
To solve the equation \(x = 2 \times \sqrt{x-1}\), first isolate the square root by dividing both sides by 2. This gives us \(\sqrt{x-1} = \frac{x}{2}\).
4Step 4: Square Both Sides
To eliminate the square root, square both sides of the equation: \((\sqrt{x-1})^2 = \left(\frac{x}{2}\right)^2\). Simplifying this gives \(x-1 = \frac{x^2}{4}\).
5Step 5: Rearrange and Simplify
Rearrange the equation: \(x - 1 = \frac{x^2}{4}\). Multiply through by 4 to clear the fraction: \(4x - 4 = x^2\). This simplifies to \(x^2 - 4x + 4 = 0\).
6Step 6: Solve the Quadratic Equation
The equation \(x^2 - 4x + 4 = 0\) is a quadratic equation that can be factored as \((x-2)^2 = 0\). Solving this gives \(x = 2\).
7Step 7: Calculate the Rectangle Dimensions
Substitute \(x = 2\) back into the expression for length to find the dimensions: \(\text{length} = \sqrt{x-1} = \sqrt{2-1} = 1\). Hence, the dimensions of the rectangle are width = 2 and length = 1.
Key Concepts
Quadratic EquationsRectangular DimensionsSquare Root
Quadratic Equations
Quadratic equations are quite significant in algebra, and their general form is \(ax^2 + bx + c = 0\). In our problem, we derived the quadratic equation \(x^2 - 4x + 4 = 0\) from the condition that the width was twice the length of a rectangle. The equation represents a parabola, and solving it involves finding the values of \(x\) that make the equation true. In this case, it's a perfect square trinomial, \((x-2)^2 = 0\), indicating that there is only one real solution: \(x = 2\). Solving quadratic equations can be done using multiple methods:
- Factoring: Break down the equation into simpler expressions. For example, \((x-2)^2 = 0\) easily shows \(x = 2\).
- Quadratic Formula: This formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), works for any quadratic equation.
- Completing the Square: A method to transform the equation into a perfect square form.
Rectangular Dimensions
Rectangular dimensions refer to the width and length of a rectangle, crucial for finding properties like area and perimeter. In this exercise, we examined a rectangle with specific conditions. The width \(x\) was given, while the length was expressed as \(\sqrt{x-1}\). This forms a direct relationship where knowing one dimension can help determine the other. Some key points about rectangles include:
- The Width is the shorter side or base of the rectangle, often denoted as \(b\).
- The Length is the longer side, which can be expressed in terms of width or other terms like \(\sqrt{x-1}\) in complex problems.
- Area is calculated using \(Area = width \times length\), a fundamental formula in geometry.
- Perimeter involves the sum of all sides and is given by \(P = 2(width + length)\).
Square Root
The square root is a critical concept in algebra that involves finding a number which, when multiplied by itself, yields the original number. In mathematical terms, if \(\sqrt{n} = a\), then \(a^2 = n\). In the exercise, the length of the rectangle involved a square root expression, \(\sqrt{x-1}\), which adds complexity to the equation.Square roots have the following properties:
- It is the inverse operation of squaring - cancelling the effect of squaring a number.
- For positive numbers, the square root is always a positive number or zero.
- Square roots can be simplified if the number is a perfect square, like \(16\) or \(9\), which have square roots \(4\) and \(3\), respectively.
Other exercises in this chapter
Problem 39
The lengths of the legs of a right triangle are 8 centimeters and 12 centimeters. Express the length of the hypotenuse in simplest radical form.
View solution Problem 40
In \(3-41\) , express each product in simplest form. Variables in the radicand with an even index are non-negative. $$ (1-\sqrt{7})(1+\sqrt{7})(1+\sqrt{7}) $$
View solution Problem 40
In \(39-42,\) find the set of real numbers for which the given radical is a real number. $$ \sqrt{9-3 x} $$
View solution Problem 40
In \(39-42,\) solve and check each equation. $$ 12 y+\sqrt{32}=\sqrt{200} $$
View solution