Question

the product of two consecutive positive integers is 812. what is the value of the lesser integer? 7 28 58 203

Explanation:

Step1: Define variables for integers

Let the lesser positive integer be \( n \), then the next consecutive integer is \( n + 1 \).

Step2: Set up the equation

The product of these two integers is 812, so we have the equation \( n(n + 1)=812 \).
Expanding this gives \( n^{2}+n - 812 = 0 \).

Step3: Solve the quadratic equation

We can try factoring or use the quadratic formula. Let's try factoring. We need two numbers that multiply to - 812 and add up to 1.
We can also test the given options:

• For \( n = 7 \): \( 7\times(7 + 1)=7\times8 = 56

eq812 \)

• For \( n = 28 \): \( 28\times(28 + 1)=28\times29=812 \)
• For \( n = 58 \): \( 58\times(58 + 1)=58\times59 = 3422

eq812 \)

• For \( n = 203 \): \( 203\times(203+ 1)=203\times204=41412

eq812 \)

Answer:

B. 28 (assuming the options are labeled as A. 7, B. 28, C. 58, D. 203)