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Let \(f(x) = x^2 + bx + c\), where \(bc \ne 0\).
If \(f(p) = f(p+3) = f(2p - 1) = 0\), which of the following could be the value of \(b\)?
I. \(3\)
II. \(-6\)
III. \(-11\)
(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
If \(f(p) = f(p+3) = f(2p - 1) = 0\), which of the following could be the value of \(b\)?
I. \(3\)
II. \(-6\)
III. \(-11\)
(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
![Let f(x) = x^2 + bx + c, where [m]bc \ne 0[/m]. If [](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "Let f(x) = x^2 + bx + c, where [m]bc \ne 0[/m]. If ["){kind=icon}
19 Apr 2026, 06:01
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Key Insight
f(x) = x2 + bx + c is a quadratic, so it can have at most 2 roots. But we're told f(p) = f(p+3) = f(2p-1) = 0 — that's 3 values giving zero.
This means at least two of these three must be equal (since a quadratic can't have 3 distinct roots).
Step 1: Set pairs equal and solve for p
Case 1: p = p + 3 → impossible (0 = 3)
Case 2: p = 2p − 1
→ p = 1
→ roots are: p = 1, p+3 = 4, 2p−1 = 1
→ So roots are 1 and 4
Case 3: p + 3 = 2p − 1
→ p = 4
→ roots are: p = 4, p+3 = 7, 2p−1 = 7
→ So roots are 4 and 7
Step 2: Find b for each case
For a monic quadratic x2 + bx + c with roots r1 and r2:
b = −(r1 + r2)
Case 2 → roots 1 and 4:
b = −(1 + 4) = −5
Case 3 → roots 4 and 7:
b = −(4 + 7) = −11 ✓
f(x) = x2 + bx + c is a quadratic, so it can have at most 2 roots. But we're told f(p) = f(p+3) = f(2p-1) = 0 — that's 3 values giving zero.
This means at least two of these three must be equal (since a quadratic can't have 3 distinct roots).
Step 1: Set pairs equal and solve for p
Case 1: p = p + 3 → impossible (0 = 3)
Case 2: p = 2p − 1
→ p = 1
→ roots are: p = 1, p+3 = 4, 2p−1 = 1
→ So roots are 1 and 4
Case 3: p + 3 = 2p − 1
→ p = 4
→ roots are: p = 4, p+3 = 7, 2p−1 = 7
→ So roots are 4 and 7
Step 2: Find b for each case
For a monic quadratic x2 + bx + c with roots r1 and r2:
b = −(r1 + r2)
Case 2 → roots 1 and 4:
b = −(1 + 4) = −5
Case 3 → roots 4 and 7:
b = −(4 + 7) = −11 ✓
