Author 
Message 
TAGS:

Hide Tags

eGMAT Representative
Joined: 04 Jan 2015
Posts: 3134

How many nonnegative integers satisfy the inequality x^2 – 7x – 18 .
[#permalink]
Show Tags
16 Jan 2019, 02:14
Question Stats:
48% (02:03) correct 52% (02:11) wrong based on 86 sessions
HideShow timer Statistics
How many nonnegative 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Manager
Joined: 01 May 2017
Posts: 82
Location: India

How many nonnegative integers satisfy the inequality x^2 – 7x – 18 .
[#permalink]
Show Tags
Updated on: 18 Jan 2019, 08:31
\((x2–7x–18)(x+5)/(x2–4)≤0\)
= \(((x+2)(x9)(x+5))/((x2)(x+2))\) = ((x9)(x+5))/(x2)) nonnegative 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, 02:25.
Last edited by Dare Devil on 18 Jan 2019, 08:31, edited 1 time in total.



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5261
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

How many nonnegative integers satisfy the inequality x^2 – 7x – 18 .
[#permalink]
Show Tags
16 Jan 2019, 02:47
simplify relation we get (x9) * ( x+5)/ ( x2) <=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 nonnegative 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



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3134

Re: How many nonnegative integers satisfy the inequality x^2 – 7x – 18 .
[#permalink]
Show Tags
18 Jan 2019, 01:48
Solution Given:• An inequality, \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\) To find:• The number of nonnegative 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 nonnegative 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
_________________



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5261
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: How many nonnegative integers satisfy the inequality x^2 – 7x – 18 .
[#permalink]
Show Tags
18 Jan 2019, 08:13
EgmatQuantExpert wrote: Solution Given:• An inequality, \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\) To find:• The number of nonnegative 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 nonnegative 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 ..




Re: How many nonnegative integers satisfy the inequality x^2 – 7x – 18 .
[#permalink]
18 Jan 2019, 08:13






