Let p and q be the roots of the quadratic equation x^2 − 15x + r = 0, and suppose that the difference between the roots is p − q = 1. What is the value of the constant r?

Introduction / Context:
This problem uses standard properties of quadratic equations, in particular the relationships between the coefficients of a quadratic and the sum and product of its roots. The equation x^2 − 15x + r = 0 has roots p and q. We are told that p − q = 1 and must find r. Instead of solving the quadratic explicitly for p and q, we use simple identities involving p + q, pq and the square of the difference between the roots. Such questions appear frequently in competitive exams to test algebraic reasoning.

Given Data / Assumptions:

• The quadratic equation is x^2 − 15x + r = 0.
• Its roots are p and q.
• p − q = 1.
• Coefficients are real, and the usual relationships between roots and coefficients hold.
• We must determine the real value of r.

Concept / Approach:
For a quadratic equation x^2 − Sx + P = 0, the sum of the roots is S and the product is P. Here, p + q = 15 and pq = r. We are given the difference p − q = 1. There is a standard identity relating the difference of roots to the sum and product: (p − q)^2 = (p + q)^2 − 4pq. Substituting known values for p + q and p − q gives a simple equation in pq = r, from which we can solve for r directly.

Step-by-Step Solution:
Step 1: From the quadratic x^2 − 15x + r = 0, identify that p + q = 15 and pq = r.Step 2: We are given that p − q = 1.Step 3: Use the identity (p − q)^2 = (p + q)^2 − 4pq.Step 4: Substitute p − q = 1, p + q = 15 and pq = r.Step 5: The equation becomes 1^2 = 15^2 − 4r.Step 6: Compute 15^2 = 225, so 1 = 225 − 4r.Step 7: Rearrange to solve for r: 4r = 225 − 1 = 224.Step 8: Therefore r = 224/4 = 56.

Verification / Alternative check:
To verify, consider the quadratic x^2 − 15x + 56 = 0. Factor this equation: 56 factors as 7 × 8, and 7 + 8 = 15, so x^2 − 15x + 56 = (x − 7)(x − 8). Hence the roots are p = 7 and q = 8 (or 8 and 7). Their difference is |p − q| = 1, as required. This confirms that r = 56 satisfies all conditions given in the problem.

Why Other Options Are Wrong:
The values 55, 60, 64 and 65 do not produce a quadratic with roots differing by 1. For example, if r were 60, the quadratic x^2 − 15x + 60 = 0 factors as (x − 5)(x − 12), whose roots differ by 7, not 1. Similar checks show discrepancies for the other options. Only r = 56 leads to factors (x − 7)(x − 8), which produce the required unit difference between roots.

Common Pitfalls:
Common mistakes include trying to solve the quadratic using the quadratic formula for each candidate r or guessing roots instead of using the root sum and product relations. Another error is misapplying the identity for (p − q)^2 or forgetting to square the given difference correctly. Using the standard relationships p + q = 15 and pq = r together with (p − q)^2 = (p + q)^2 − 4pq is the most reliable and efficient approach.

Final Answer:
The required constant in the quadratic equation is r = 56.