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13 May 2024, 01:30
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Difficulty:
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Question Stats:
59% (01:49) correct
41%
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wrong
based on 292
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If \(x\) is an integer and \(2x^2+9 < 9x\), how many positive factors does \(x\) have?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
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01 Jun 2024, 08:30
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Official Solution:
If \(x\) is an integer and \(2x^2+9 < 9x\), how many positive factors does \(x\) have?
A. 1
B. 2
C. 3
D. 4
E. 5
First, rearrange the inequality: \(2x^2 - 9x + 9 < 0\). Now, factor the quadratic: \((2x - 3)(x - 3) < 0\). The roots are \(\frac{3}{2}\) and 3. Since the inequality sign is "\( < \)", the solution lies between the roots: \(1.5 < x < 3\). The only integer in this range is 2, so \(x=2\), which has two positive factors 1 and 2.
Answer: B
If \(x\) is an integer and \(2x^2+9 < 9x\), how many positive factors does \(x\) have?
A. 1
B. 2
C. 3
D. 4
E. 5
First, rearrange the inequality: \(2x^2 - 9x + 9 < 0\). Now, factor the quadratic: \((2x - 3)(x - 3) < 0\). The roots are \(\frac{3}{2}\) and 3. Since the inequality sign is "\( < \)", the solution lies between the roots: \(1.5 < x < 3\). The only integer in this range is 2, so \(x=2\), which has two positive factors 1 and 2.
Answer: B
General Discussion
13 May 2024, 01:55
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Bunuel
\(2x^2+9 < 9x\)
\(2x^2+ 9 - 9x< 0 \)
The roots of the equation are \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)\(\)
Roots are ⇒ \(\frac{9 \pm \sqrt{81 - 72}}{4} \)
\(\frac{9 \pm \sqrt{9}}{4} \)
Roots are ⇒ \(\frac{9 + 3}{4} = 3 \) , \(\frac{9 - 3}{4} = 1.5 \)
Hence, \(x\) lies between \(1.5\) and \(3\). As x is an integer, the only possible value of x = \(2\)
2 has two positive factors.
Option B
