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If h and k are both positive integers, are both roots of the equation 2x^2 + hx + k = 0 integers?
(1) The product of the roots is 6
(2) kh = 132
Source: Knewton
Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.
Thanks.
(1) The product of the roots is 6
(2) kh = 132
Source: Knewton
Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.
Thanks.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
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justbequiet
Joined: 04 Sep 2012
Last visit: 15 Apr 2016
Posts: 89
Location: Philippines
Concentration: Marketing, Entrepreneurship
Schools: Ross (Michigan) - Class of 2017
GMAT 1: 620 Q48 V27
GMAT 2: 660 Q47 V34
GMAT 3: 700 Q47 V38
GPA: 3.25
WE:Sales (Manufacturing)
hubahuba
(1) Insufficient
\((2x+3)(x+4) = \frac{3}{2}*4=6\) One root is an integer, the other is not.
\((2x+6)(x+2) = \frac{6}{2}*2=6\) Both roots are Integers
(2) Sufficient
Prime Roots of \(132 = 2^2*3*11\)
Only 2 combinations are possible
If h=-3 & k=-44
\((2x+11)(x-4)\)--> One root is an integer, the other is not.
If h=11 & k=12
\((2x+3)(x+4)\) --> One root is an integer, the other is not.
Answer is B
hubahuba
Let the roots of the equation be P and Q.
p+q = -h/2--(i) & pq = k/2 --(ii).
Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.
Statement 2 -- Given kh =132
132 = 132 * 1
= 66 * 2
= 44 * 3
= 33 * 4
= 22 * 6
= 11 * 12
Thus these are the possible values k or h can take since given they are positive integers.
k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.
Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values.
Thus a definite solution is obtained.
Statement 2 alone is sufficient
Posts: 138
justbequiet
Why only two combinations are possible ? Why not 22,6 or 66,2
justbequiet
Joined: 04 Sep 2012
Last visit: 15 Apr 2016
Posts: 89
Location: Philippines
Concentration: Marketing, Entrepreneurship
Schools: Ross (Michigan) - Class of 2017
GMAT 1: 620 Q48 V27
GMAT 2: 660 Q47 V34
GMAT 3: 700 Q47 V38
GPA: 3.25
WE:Sales (Manufacturing)
himanshujovi
Its something I overlooked when trying out the options. But regardless, result will be the same.
VenoMftw
Thanks for such a simple solution. I completely missed using the integer condition for h and k. +1 Kudos for you.
VenoMftw
When you reach this point, how do you know that solving will give vales that won't be integers?
Did you actually try to solve or do you have a shortcut?
VenoMftw
I couldn't understand this line k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.
Where does the question say that the sum and product of the roots are integers.
prasun9
It is not given that sum and product need to be integers. The logic of this question is a little more twisted:
Question: Are both roots positive?
Answer: Yes/No Or May be
If we answer with a certain Yes or a certain No, our stmnt is sufficient.
If we answer with May be (sometimes Yes, sometimes No), our statement is not sufficient.
2x^2 + hx + k = 0
Sum of roots = -h/2
Product of roots = k/2
Let's find all those cases in which we can answer with 'Yes'
Stmnt 1: The product of the roots is 6.
k/2 = 6 so k = 12
The roots could be -2, -3 such that the equation would be x^2 + 5x + 6 = 0 or 2x^2 + 10x + 12 = 0
The roots could be no integer for any random value of h.
So we have no definite answer from this statement. Not sufficient.
Stmnt 2: kh = 132
Let's see whether given this information, we can find both integer roots.
kh = 132 = 2^2 * 3 * 11
If both roots are integers, sum and product of roots must be integers (k/2 and h/2 must be integers)
So k and h must be even. So give a 2 to each of k and h and split 3 and 11 in two ways (no need to worry about negative sign):
k and h could be 2 and 66 such that product and sum will be 1 and 33 - Can the product/sum of two integers be 33 while sum/product is 1? No.
k and h could be 6 and 22 such that product and sum will be 3 and 11 - Can the product/sum of two integers be 11 while sum/product is 3? No.
So it is not possible to have both integers. We can answer with a definite 'No'.
Hence this statement alone is sufficient.
Answer (B)
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VeritasPrepKarishma
Please, how did you arrive at the "sum of roots" and "product of roots"? I am totally lost on this part...
Jada
This is a standard formula for the roots of quadratic equations. Assume the quadratic equation is ax^2 + bx + c = 0
Sum of roots = -b/a
Product of roots = c/a
Jada
Check out the blog posts on the link given in my signature below.
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SZolotukhin
Joined: 23 May 2015
Last visit: 04 Aug 2022
Posts: 2
Location: Ukraine
GMAT 1: 720 Q50 V38
GPA: 3.99
WE:Analyst (Finance: Venture Capital)
GMAT 1: 720 Q50 V38
Posts: 2
Hey, basically I came to the correct answer B but in other way:
(1) x1*x2=6=k/2
k=12
2x^2+hx+12=0
Basicaly, roots might be integers or non-integers so insufficient.
(2) h*k=132 (I)
x1+x2=-h/2 (II)
x1*x2=k/2 (III)
then I multiply (II) * (III) and get
x1*x2*(x1+x2)=-k*h/4=-132/4=-33
So, if we asssume that x1 and x2 integers, what we should conclude is INTEGER1*INTEGER2*INTEGER3=-33 (which is possible only for triplet 1/3/11 with one/all of them negative - clearly among 1/3/11 there are no pairs to satisfy a+b=c =>> therefore, x1 OR x2 should be non-integer). So (2) is sufficient.
Not sure is it correct reasoning, but the answer is OK.
(1) x1*x2=6=k/2
k=12
2x^2+hx+12=0
Basicaly, roots might be integers or non-integers so insufficient.
(2) h*k=132 (I)
x1+x2=-h/2 (II)
x1*x2=k/2 (III)
then I multiply (II) * (III) and get
x1*x2*(x1+x2)=-k*h/4=-132/4=-33
So, if we asssume that x1 and x2 integers, what we should conclude is INTEGER1*INTEGER2*INTEGER3=-33 (which is possible only for triplet 1/3/11 with one/all of them negative - clearly among 1/3/11 there are no pairs to satisfy a+b=c =>> therefore, x1 OR x2 should be non-integer). So (2) is sufficient.
Not sure is it correct reasoning, but the answer is OK.
hubahuba
here is my take
2x^2 + hx + k = 0 can be expressed as x^2 + h/2 x + k/2 = 0 where , -h/2 = the sum of the roots and k/2 is the product.
from 1
k/2 = 6 , thus the roots ( x,y) product is 6 , xy = 6 = 3*2 or sqrt 3 * 2sqrt 3 ... insuff
from2
kh = 132 thus kh/4 = 3*11 , assuming roots are x,y thus xy(x+y) = 3*11 thus x+y = 3*11/xy for ( x+y) to be integer xy must either 33 , -33 , 1 or -1
but if xy = 3*11 or -3*11 then (x+y) = 1 or -1 ( impossible)
B
gauravk
Say h = 11
Sum of the roots = -h/2 = -11/2 = -5.5
Product of the roots = 6
The roots would be -1.5 and -4.
Both roots may not be integers.
So the first statement is not sufficient.
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hubahuba
why h & k cannot be odd? Product and sum of the roots can have decimal values.
VenoMfTw
What prevents h/2 and k/2 from having decimal values. It is not mentioned anywhere in the question or statements.
Posts: 12
VeritasKarishma
Hey, can you explain how did you arrive at Sum of roots = -h/2 Product of roots = k/2 ?
menonpriyanka92
Given a quadratic equation, \(ax^2 + bx + c = 0\)
Sum of roots = -b/a
Product of roots = c/a
You can easily derive them by expanding the following:
\(ax^2 + bx + c = a(x – p)(x – q)\) (where p and q are the roots of the equation)
\(ax^2 + bx + c = ax^2 -a(p + q)x + apq\)
\(b = - a(p + q)\)
\((p + q) = -b/a\)
\(c = apq\)
\(pq = c/a\)
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