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If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 18:02
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GMATBusters’ Quant Quiz Question 1
For past quiz questions, click here If \(x^2 + kx  72 = 0\) has integer roots, how many integer values can k take? A. 6 B. 8 C. 9 D. 10 E. 12
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 18:09
As it is similar to ax2 + bx + c, the product of roots = c/a, here c/a= 72. If, 72 can be written as product of 2 integers, e.g. x and y, then sum of roots = x+y = b/a = k
If (x,y) = (72,1), then x+y (k) = 71, If (x,y) = (72,1), then x+y = 71,
Similarly, (x,y) can be (36,2),(36,2),(24,3),(24,3),(18,4),(18,4),(12,6),(12,6),(9,8),(9,8) So, total is 12. Imo. E



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 18:51
\(x^2 (sum of roots)x+ products of roots= 0\) product of roots= 72 That means one of the two roots is negative. Also k has to be negative since the equation quadratic becomes \(x^2+kx72=0\) If we factor 72 to check for the possibilities, we get 6 possibilities
9 and +8 18 and +4 36 and +2 72 and +1 24 and +3 12 and +6
Answer: A



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 19:19
K= 72+1=71 36+2=34 24+3=21 18+4=14 12+6=6 9+8 =1 Also the reverse of this 721=71 362=34 243=21 184=14 126=6 98 =1 So total 12 values of k are possible. Option E is the answer
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 19:46
ANSWER:E x2+kx72=0 x(x+k)=72 Following combinations can satisfy the condition given (72,1)(1,72)(36,2)(2,36)(3,24)(24,3)(4,18)(18,4)(6,12)(12,6)(8,9)(9,8) Hence total value of k=12



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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Updated on: 07 Jun 2020, 06:20
sum of roots k/1 and product ; 72 since sum of roots is ve & so is the product so either one of them has to be +ve and other total factors of 72 ; 2^3*3^2 ; 4*3 ; 12 OPTION E
If x2 + kx  72 = 0 has integer roots, how many integer values can k take?
A. 6 B. 8 C. 9 D. 10 E. 12
Originally posted by Archit3110 on 06 Jun 2020, 19:55.
Last edited by Archit3110 on 07 Jun 2020, 06:20, edited 1 time in total.



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 20:29
A Positive & negative of 1, 6 & 14
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 21:46
as this equation has integer roots, it can be expressed as (x+a) (x+b) where ab =72 now after factorization it is understood that 72 can be expressed as a multiple of two numbers as follows: 1,72 2,36 3,24 4,18 6,12 8,9
For product to be 72, either of the numbers for each pair has to be negative, which means for each pair above, (a+b) shall have two value therefore, the answer is 6x2= 12



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 22:08
Asked: If x2+ kx  72 = 0 has integer roots, how many integer values can k take? 72 = 2*2*2*3*3 = 2^3*3^2 k should be difference between factors and should be integer k = 72  1 = 71; Case 1: k = 71: x^2 + 71x  72 = x^2 + 72x  x  72 = (x1)(x+72) = 0 ; x = 1 or 72 Case 2: k =  71: x^2 71x 72 = x^2 72x + x  72 = (x+1)(x72) = 0 ; x = 1 or 72 k = 36  2 = 34 k = 34 or  34 k = 18  4 = 14 k = 14 or  14 k = 9  8 = 1 k = 1 or  1 k = 24  3 = 21 k = 21 or 21 k = 12  6 = 6 k = 6 or  6 Total values that k can take = 6*2 = 12 A. 6 B. 8 C. 9 D. 10 E. 12 IMO E
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 22:17
x2 + kx  72 = 0 has integer roots, how many integer values can k take?
A. 6 B. 8 C. 9 D. 10 E. 12
Solution:
For the quadratic equation \(x^2 + kx  72 = 0 \) the product of the roots is 72. ( c/a is the product of roots for \(ax^2 + bx + c = 0\))
Therefore, the number of integer roots will be the number of ways 72 can be expressed as the product of two numbers. So we need to find the number of factors of 72.
The prime factorization of 72 is \(2^2 * 3^2 \)
The total number of factors are (3+1)*(2+1) = 12
Hence E is the correct answer



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 22:34
IMO E
X^2+KX72 = 0 Factorize 72 = 2^3*3^2 Total no. of factors = (3+1)(2+1) => 4*3 => 12
Roots of equation x^2+kx72 = 0 Product of roots = c/a = 72/1 = 72 => (72*1),(72*1),(36*2),(36*2),(24*3),(24*3),(18*4),(18*4),(12*6),(12*6),(9*8),(9*8) => roots. Sum of roots = b/a = k 01) => 72+1 = 71 02) => +721 = +71 03) => 36+2 = 34 04) => +362 = +34 05) => 24+3 = 21 06) => +243 = +21 07) => 18+4 = 14 08) => +184 = +14 09) => 12+6 = 06 10) => +126 = +06 11) => 09+8 = 01 12) => +098 = +01 K can take 12 integer values
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 23:24
Quote: If x^2 + kx  72 = 0 has integer roots, how many integer values can k take?
A. 6 B. 8 C. 9 D. 10 E. 12 E, imo x^2 + kx  72 = 0 so x (x+k) = 72= 2^3 * 3^2 so total factors of 72 = 4 * 3 = 12 so we will have 12 possible value of k ,which will satisfy the condition .
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 23:41
\(x^{2}+kx72=0\) Find number of factors for 72: 72= \(2^{3}*3^{2}\)Number of factors = (a+1)(b+1)..... where a and b are exponents of the prime factors.Number of factors = 4*3 = 12 ............. (1,2,3,4,6,8,9,12,18,24,36,72)The number of pairs of factors that make 72 = 12/2 = 6 pairs............ (1*72, 2*36...........8*9)Since, the value is 72, one factor must be positive while the other negative. For ex the pair 1x72 can we written in two ways: \((i) 1*(72)\) \((ii) (1)*72\)Since k is sum of the factors, the above two pairs lead two different sums \((i) 1 72 = 71\) \((ii) 1+ 72 = 71\) Thus, a single pair of factors gives 2 values of k. 6 pairs will give 12 values of k. (\(\pm71, \pm34, \pm21, \pm14, \pm6, \pm1\)) +1 kudos if you like the solution



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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06 Jun 2020, 23:49
x^2 + kx  72 = 0.
The Product of the roots: 72. The sum of roots : k Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. If one no. in the pair is +ve other would be ve. So we can have a total of 6 pairs of roots.
Answer A.



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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07 Jun 2020, 07:37
x^2 +kx72=0,
For any quadratic equation, sum of the roots= b/a, in this case = k/1 = k product of the roots = c/a, in this case = 72/1 = 72 Essentially, we have to find the prime factors of 72. 72= 3^2 x 2^3 No of factors = (2+1)(3+1) = 12. Product of any two factors will yield the value of 72 and the summation/difference of the same two factors will give me the value of K.
Answer E.



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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07 Jun 2020, 07:54
The method I used here was finding possible combinations of integers with their multiplication = 72. If we do so, we will get the following: 1,72 2,36 3,24 4,18 6,12 8,9 9,8 12,6 18,4 24,3 36,2 72,1 Hence 12 combinations.
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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07 Jun 2020, 10:15
We list pairs of factors of 72 which can satisfy the equation:
1. 72*1 2. 36*2 3. 12*6 4. 9*8 5. 4*18 6. 3*24
Now each of these pairs can be written in 2 different ways. Eg x^212x +6x72 and x^2 6x+12x72 So there are 12 k values possible.
Hence answer is E.
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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07 Jun 2020, 10:59
If x2+kx−72=0x2+kx−72=0 has integer roots, how many integer values can k take?
A. 6 B. 8 C. 9 D. 10 E. 12 x(x+k)=72, 72 can be expresses as (1*72),(2*36),(3*24),(4*18),(6*12),(8*9). k can be either the first factor or when x is the first factor, k will be the second factor minus x. So total 12 integer values.



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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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07 Jun 2020, 10:59
First, we need to satisfy the condition that the roots are integers so we analyze the determinant: \(b^24ac>=0\) \(k^2(4)(1)(72)>=0\) once we reached this expression, it is clear that k can take any real value to satisfy the inequality Then, we know that the original expression can be factorized this form: \((x+a)(x+b)=0\) where a and b are the integer roots The coefficient of the linear term (k) will be given by all the possible results of the arithmetic addition of a and b Now, we count the number of factors of 72 \(72 = 2^3.3^2\) # of factors of \(72 = (3+1)(2+1) = 12\) Then, we will have 6 combinations of two factors whose product is 72, but because the independent term is negative (72), we will have a total of 12 possible combinations To illustrate: 1 and 72, so the expression will become \((x1)(x+72)\) and k = 71 1 and 72, so the expression will become \((x+1)(x72)\) and k = 71 There are 12 such couples, so the answer to the exercise is E) 12
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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07 Jun 2020, 20:44
Need 2 integer values s.t : Product of those 2 integer values is 72 Sum of those 2 integer values is k Product______________________________ Sum(K) (1, 72)_________________________________  71 (1, 72)__________________________________+71 (2, 36)__________________________________34 (2,36)___________________________________+34 (3, 24)___________________________________21 (3, 24)___________________________________+21 (4, 18)___________________________________14 (4,18)____________________________________+14 (6,12)____________________________________6 (6,12)____________________________________+6 (8,9)_____________________________________1 (8,9)_____________________________________+1 Thus possible integer values for k=12 Answer:E
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Re: If x^2 + kx  72 = 0 has integer roots, how many integer values can k
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