If 3x + 7, x^2 + p, and 7x + 5 all represent the same value for some real number x (that is, 3x + 7 = x^2 + p = 7x + 5), what is the value of p?

Introduction:
This problem tests solving linked equalities. When multiple expressions are stated to be equal to the same value, you can equate them pairwise. The quickest path is to first solve for x by equating the two linear expressions (3x + 7 and 7x + 5). Once x is known, plug into x^2 + p = the common value to find p. This approach avoids unnecessary algebra with p until the end and keeps the steps clean.

Given Data / Assumptions:

• 3x + 7 = x^2 + p = 7x + 5
• x is a real number
• We need to find p

Concept / Approach:
Since 3x + 7 and 7x + 5 are both equal to the same quantity, they must be equal to each other. Solve 3x + 7 = 7x + 5 to get x. Then use x^2 + p = 3x + 7 (or = 7x + 5) to compute p. Finally express p as a fraction to match the options exactly.

Step-by-Step Solution:
Set the linear expressions equal: 3x + 7 = 7x + 5 Bring terms together: 7 - 5 = 7x - 3x 2 = 4x x = 1/2 Common value = 3x + 7 = 3*(1/2) + 7 = 3/2 + 7 = 17/2 Now x^2 + p = 17/2 x^2 = (1/2)^2 = 1/4 p = 17/2 - 1/4 = (34/4 - 1/4) = 33/4

Verification / Alternative check:
Check with 7x + 5: for x = 1/2, 7*(1/2)+5 = 7/2 + 5 = 17/2, same common value. Then x^2 + p = 1/4 + 33/4 = 34/4 = 17/2, consistent. So p is correct.

Why Other Options Are Wrong:
17/2 is the common value, not p. The other fractions do not satisfy x^2 + p = 17/2 when x = 1/2. Only 33/4 makes the equality hold.

Common Pitfalls:
Confusing p with the common value, forgetting to square x correctly, or subtracting fractions incorrectly when computing p.

Final Answer:
The value of p is 33/4.