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Difficulty:
95%
(hard)
Question Stats:
27% (02:46) correct
73%
(03:08)
wrong
based on 22
sessions
History
Date
Time
Result
Not Attempted Yet
The equation \(x^2 -6x + k = 100\) has two distinct integer solutions. If the product of the solutions is negative, how many possible integer values of \(k\) are positive ?
A) 6
B) 7
C) 8
D) 9
E) 10
A) 6
B) 7
C) 8
D) 9
E) 10
03 May 2026, 23:05
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Distinct integer solutions: b^2 - 4ac >0 and a perfect square.
4(109-k) => 109 -k to be a perfect square
Now product of roots is -ve => k-100<0 ----> k<100.
For k = 100. we can have 109 -k =9 which is 3 square.
So the values are possible for squares of from 10 to 4
Ans B
Hope it helps.
Hi kevincan,
It would be helpful to know if my approach is logically correct. Also, if any, I want to know other approach to this question
4(109-k) => 109 -k to be a perfect square
Now product of roots is -ve => k-100<0 ----> k<100.
For k = 100. we can have 109 -k =9 which is 3 square.
So the values are possible for squares of from 10 to 4
Ans B
Hope it helps.
Hi kevincan,
It would be helpful to know if my approach is logically correct. Also, if any, I want to know other approach to this question
kevincan
03 May 2026, 23:14
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Bookmarks
So here,
x2 − 6x + k = 100
⇒ x2 − 6x + (k − 100) = 0
So we have roots a, b
a + b = 6
a × b = (k − 100)
---
Condition given:
product is negative
⇒ (k − 100) < 0
---
So the only way to get a × b negative
is one value positive and one value negative
---
Now checking integer pairs (a, b):
(7, −1)
(8, −2)
(9, −3)
(10, −4)
(11, −5)
(12, −6)
(13, −7)
---
Shortcut thinking (no need to calculate everything):
As soon as product crosses 100 in magnitude,
k becomes negative or not valid for condition
So we only take values till product stays within range
Example:
(14, −8) → product = −112
→ k = −12 (not needed / discard)
So we stop earlier itself
---
Then checking k:
7 × (−1) = −7 → k − 100 = −7 → k = 93
(8, −2) → −16 → k = 84
(9, −3) → −27 → k = 73
(10, −4) → −40 → k = 60
(11, −5) → −55 → k = 45
(12, −6) → −72 → k = 28
(13, −7) → −91 → k = 9
---
So, total 7 values
---
Final Answer: 7
---
Hope this helps
Feel free to tag me if needed
– Rajdeep
x2 − 6x + k = 100
⇒ x2 − 6x + (k − 100) = 0
So we have roots a, b
a + b = 6
a × b = (k − 100)
---
Condition given:
product is negative
⇒ (k − 100) < 0
---
So the only way to get a × b negative
is one value positive and one value negative
---
Now checking integer pairs (a, b):
(7, −1)
(8, −2)
(9, −3)
(10, −4)
(11, −5)
(12, −6)
(13, −7)
---
Shortcut thinking (no need to calculate everything):
As soon as product crosses 100 in magnitude,
k becomes negative or not valid for condition
So we only take values till product stays within range
Example:
(14, −8) → product = −112
→ k = −12 (not needed / discard)
So we stop earlier itself
---
Then checking k:
7 × (−1) = −7 → k − 100 = −7 → k = 93
(8, −2) → −16 → k = 84
(9, −3) → −27 → k = 73
(10, −4) → −40 → k = 60
(11, −5) → −55 → k = 45
(12, −6) → −72 → k = 28
(13, −7) → −91 → k = 9
---
So, total 7 values
---
Final Answer: 7
---
Hope this helps
Feel free to tag me if needed
– Rajdeep
