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Updated on: 23 Sep 2025, 12:28
Originally posted by Sajjad1994 on 23 Sep 2025, 12:21.
Last edited by Sajjad1994 on 23 Sep 2025, 12:28, edited 1 time in total.
Last edited by Sajjad1994 on 23 Sep 2025, 12:28, edited 1 time in total.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
71% (00:45) correct
28% (01:24) wrong based on 7 sessions
History
Date
Time
Result
Not Attempted Yet
27 Sep 2025, 20:59
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(x^2 +9)(x+9)(x-5) =0
x^2 +9=0 ; x^2 = -9. ; x= (-9)^(1/2) .....not real solution
(x+9)=0 ; x= -9
(x-5)=0. ; x=+5
X can be +5 or -9
Quantity A: x = +5, -9
Quantity B: 7
7>5 & 7>-9
B>A
B
x^2 +9=0 ; x^2 = -9. ; x= (-9)^(1/2) .....not real solution
(x+9)=0 ; x= -9
(x-5)=0. ; x=+5
X can be +5 or -9
Quantity A: x = +5, -9
Quantity B: 7
7>5 & 7>-9
B>A
B
29 Sep 2025, 07:06
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Official Explanation
The product of the three quantities is zero, so one of the individual factors \((x^2 + 9), (x + 9),\) or \((x – 5)\) must equal zero. \(x^2 + 9 = 0\) has no real solution, because \(x^2\) can’t equal \(–9,\) or any negative. If \(x + 9 = 0\) then \(x = –9,\) and if \(x – 5 = 0,\) then \(x = 5.\) So the two possible values of \(x\) are \(– 9\) and 5, both less than 7. Then Quantity (B), 7, is greater.
Answer: B
The product of the three quantities is zero, so one of the individual factors \((x^2 + 9), (x + 9),\) or \((x – 5)\) must equal zero. \(x^2 + 9 = 0\) has no real solution, because \(x^2\) can’t equal \(–9,\) or any negative. If \(x + 9 = 0\) then \(x = –9,\) and if \(x – 5 = 0,\) then \(x = 5.\) So the two possible values of \(x\) are \(– 9\) and 5, both less than 7. Then Quantity (B), 7, is greater.
Answer: B
