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How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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 16 Jan 2019, 01:14
Question Stats:
49% (02:12) correct 51% (02:19) wrong based on 109 sessions
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How many non-negative integers satisfy the inequality \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)? A. 6
B. 7
C. 8
D. 9
E. Infinite To read all our articles: Must read articles to reach Q51
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How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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Updated on: 18 Jan 2019, 07:31
\((x2–7x–18)(x+5)/(x2–4)≤0\)
= \(((x+2)(x-9)(x+5))/((x-2)(x+2))\)
= ((x-9)(x+5))/(x-2))
non-negative implies 0,1,2,3,4.....
when x = 0,1 -ve sign cancels
at x= 2 denominator is zero implies the value is infinty. So, x!= 2
x = 3,4,5,6,7,8,9
numerator is negative OR Zero and denominator is positive
Rest of the values x >9 given equation is positive
So, Given equation satisfies, implies for 7 integers it satisfies the given condition
Option B is correct
Originally posted by Dare Devil on 16 Jan 2019, 01:25.
Last edited by Dare Devil on 18 Jan 2019, 07:31, edited 1 time in total.
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Joined: 18 Aug 2017
Posts: 6412
Location: India
Concentration: Sustainability, Marketing
GPA: 4
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How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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16 Jan 2019, 01:47
simplify relation we get (x-9) * ( x+5)/ ( x-2) <=0 values of x would satisfy relation <=0 x at 2,3,4,5,6,7,8,9 satisfies the relation as for 0,1, we get >0 values and for integers beyond 9 as well we get >0 values IMO C ; 8 EgmatQuantExpert wrote: How many non-negative integers satisfy the inequality \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)? A. 6
B. 7
C. 8
D. 9
E. Infinite To read all our articles: Must read articles to reach Q51 |
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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3410
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Re: How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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18 Jan 2019, 00:48
Solution Given:• An inequality, \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\) To find:• The number of non-negative integer values of x that satisfy the given inequality Approach and Working: • \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\) Let’s factorise the quadratic expressions in the numerator and denominator • \(\frac{(x – 9)(x + 2)(x + 5)}{(x + 2)(x - 2)} ≤ 0\) We can cancel out the common term, x + 2, in both numerator and denominator • \(\frac{(x – 9)(x + 5)}{(x - 2)} ≤ 0\) Now, multiplying and dividing by (x – 2), we get, • \(\frac{(x – 2)(x - 9)(x + 5)}{(x - 2)^2} ≤ 0\) In the above expression, the denominator is always greater than 0. Thus, the numerator must be ≤ 0. • (x – 2)(x - 9)(x + 5) ≤ 0 Representing the zero points {-5, 2, 9} on a number line, we can find the regions in which the expression is ≤ 0  • From the above figure, we can see that (x – 2)(x - 9)(x + 5) ≤ 0 in the regions, 2 ≤ x ≤ 9 and x ≤ -5
• And, we know that x ≠ 2, since, the denominator of the given expression cannot be = 0
We are asked to find the number of non-negative integer values of x. • Therefore, the values of x can be {3, 4, 5 ,6 ,7 ,8, 9} Hence the correct answer is Option B. Answer: B
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Joined: 18 Aug 2017
Posts: 6412
Location: India
Concentration: Sustainability, Marketing
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Re: How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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18 Jan 2019, 07:13
EgmatQuantExpert wrote: Solution Given:• An inequality, \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\) To find:• The number of non-negative integer values of x that satisfy the given inequality Approach and Working: • \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\) Let’s factorise the quadratic expressions in the numerator and denominator • \(\frac{(x – 9)(x + 2)(x + 5)}{(x + 2)(x - 2)} ≤ 0\) We can cancel out the common term, x + 2, in both numerator and denominator • \(\frac{(x – 9)(x + 5)}{(x - 2)} ≤ 0\) Now, multiplying and dividing by (x – 2), we get, • \(\frac{(x – 2)(x - 9)(x + 5)}{(x - 2)^2} ≤ 0\) In the above expression, the denominator is always greater than 0. Thus, the numerator must be ≤ 0. • (x – 2)(x - 9)(x + 5) ≤ 0 Representing the zero points {-5, 2, 9} on a number line, we can find the regions in which the expression is ≤ 0 • From the above figure, we can see that (x – 2)(x - 9)(x + 5) ≤ 0 in the regions, 2 ≤ x ≤ 9 and x ≤ -5
• And, we know that x ≠ 2, since, the denominator of the given expression cannot be = 0
We are asked to find the number of non-negative integer values of x. • Therefore, the values of x can be {3, 4, 5 ,6 ,7 ,8, 9} Hence the correct answer is Option B. Answer: B@EgmatQuantExpert" the expression [m]\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0 so wont it be =0 and should be valid at x=2 would be a valid to count as an integer ... considering this relation only I opted for option C '8' over 7 option b ..
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Re: How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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29 Mar 2020, 03:14
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Re: How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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29 Mar 2020, 03:14
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