Two consecutive odd whole numbers have a product of 483 . What is the smaller of the two numbers?

To find the two consecutive odd whole numbers whose product is 483, let's call the smaller number \( x \). The next consecutive odd number would then be \( x + 2 \). We can set up the equation: \[ x(x + 2) = 483 \] Expanding this gives: \[ x^2 + 2x - 483 = 0 \] Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = 2, c = -483 \). Plugging in the values: \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-483)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 1932}}{2} = \frac{-2 \pm \sqrt{1936}}{2} \] Since \( \sqrt{1936} = 44 \): \[ x = \frac{-2 \pm 44}{2} \] Calculating the two possible solutions: 1. \( x = \frac{42}{2} = 21 \) 2. \( x = \frac{-46}{2} = -23 \) (which we discard as we need whole numbers) So the smaller of the two consecutive odd whole numbers is \( 21 \). Let’s check: The two numbers are \( 21 \) and \( 23 \), and their product \( 21 \times 23 \) indeed equals \( 483 \). The smaller number is **21**.