GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Oct 2018, 15:07

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50009
y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

09 Mar 2016, 12:57
2
14
00:00

Difficulty:

95% (hard)

Question Stats:

35% (02:17) correct 65% (02:20) wrong based on 232 sessions

### HideShow timer Statistics

y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k are integers, what is the least value of b?

A. 0
B. –32
C. –255
D. –256
E. –257

_________________
Manager
Joined: 09 Jul 2013
Posts: 109
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

10 Mar 2016, 14:56
4
1
If we were to factor $$x^2+bx+256 =0$$ we would get

$$(x-h)(x-k)=0$$, where h and k are the roots of the equation (as given - "...cuts the x axis at (h, 0) and (k, 0)").

So (-h)*(-k) = 256 and -h-k = b

Possible combinations for h and k are the factor pairs of 256:
1, 256
2, 128
4, 64
etc...

And also the negative of these pairs.
-1, -256
-2, -128
etc.

We can see that to have the smallest value of h+k, they must be -1 and -256

Therefore the lowest value of b = -1-256 = -257

_________________

Dave de Koos

##### General Discussion
Intern
Joined: 22 Mar 2015
Posts: 31
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

10 Mar 2016, 05:15
Bunuel wrote:
y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k are integers, what is the least value of b?

A. 0
B. –32
C. –255
D. –256
E. –257

Since value of a is positive, parabola opens in upward direction.Therefore,
x co-ordinate of parabola vertex = -b/2a
(h+k)/2=-b/2a
a=1 so,
h+k=-b so,
b=-(h+k)
IMO Option E
Intern
Joined: 25 Feb 2016
Posts: 3
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

10 Mar 2016, 22:53
1
As the curve cuts the x-axis at (h,0) and (k,0). Therefore h,k are the roots of the quadratic equation.

For the quadratic equation is in the form of ax^2+bx+c=0,

The product of the roots =c/a= 256/1=256 and the sum of the roots = -b/a=-b

256 can be expressed as product of two numbers in the following ways:
1 * 256
2 * 128
4 * 64
8 * 32
16 * 16
The sum of the roots is maximum when the roots are 1 and 256 and the maximum sum is 1 + 256 = 257.

The least value possible for b is therefore -257.

Manager
Joined: 01 Mar 2014
Posts: 117
Schools: Tepper '18
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

17 Mar 2016, 04:34
AshokGmat2016 wrote:
As the curve cuts the x-axis at (h,0) and (k,0). Therefore h,k are the roots of the quadratic equation.

For the quadratic equation is in the form of ax^2+bx+c=0,

The product of the roots =c/a= 256/1=256 and the sum of the roots = -b/a=-b

256 can be expressed as product of two numbers in the following ways:
1 * 256
2 * 128
4 * 64
8 * 32
16 * 16
The sum of the roots is maximum when the roots are 1 and 256 and the maximum sum is 1 + 256 = 257.

The least value possible for b is therefore -257.

I know this is rather a simple question but i am having trouble understanding the basic concepts. Can someone please elaborate??
Math Expert
Joined: 02 Aug 2009
Posts: 6974
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

17 Mar 2016, 05:17
MeghaP wrote:
AshokGmat2016 wrote:
As the curve cuts the x-axis at (h,0) and (k,0). Therefore h,k are the roots of the quadratic equation.

For the quadratic equation is in the form of ax^2+bx+c=0,

The product of the roots =c/a= 256/1=256 and the sum of the roots = -b/a=-b

256 can be expressed as product of two numbers in the following ways:
1 * 256
2 * 128
4 * 64
8 * 32
16 * 16
The sum of the roots is maximum when the roots are 1 and 256 and the maximum sum is 1 + 256 = 257.

The least value possible for b is therefore -257.

I know this is rather a simple question but i am having trouble understanding the basic concepts. Can someone please elaborate??

Hi,

let me try to explain it to you..

y = x^2 + bx + 256..
when we say the curve cuts the x-axis at (h,0) and (k,0), it means h and k are the roots of equation since here y is 0
x^2 + bx + 256 = 0...
We can see both (h,0) and (k,0) have y as 0, which converts the equation y = x^2 + bx + 256.. as x^2 + bx + 256 = 0...

for a quadratic equation, ax^2+bx+c=0, the product of its roots are c/a and sum= -b/a..
here a=1, b=b and c=256
so here product= hk=c=256..
sum= -b =h+k
sice we are looking for the max sum, so as to have least b, we take h and k as 256 and 1..
P=256*1=256
so SUM = -b= 256+1=257
so b=-257
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Math Expert
Joined: 02 Aug 2009
Posts: 6974
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

17 Mar 2016, 05:20
davedekoos wrote:
If we were to factor $$x^2+bx+256 =0$$ we would get

$$(x-h)(x-k)=0$$, where h and k are the roots of the equation (as given - "...cuts the x axis at (h, 0) and (k, 0)").

So (-h)*(-k) = 256 and -h-k = b

Possible combinations for h and k are the factor pairs of 256:
1, 256
2, 128
4, 64
etc...

And also the negative of these pairs.
-1, -256
-2, -128
etc.

We can see that to have the smallest value of h+k, they must be -1 and -256

Therefore the lowest value of b = -1-256 = -257

Hi,
you have gone wrong in coloured portion..
For lowest value we require MAX sum as SUM= -b/a
so h and k will be 1 and 256 and not -1 and -256
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Manager
Joined: 01 Mar 2014
Posts: 117
Schools: Tepper '18
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

17 Mar 2016, 05:27
chetan2u wrote:
MeghaP wrote:
AshokGmat2016 wrote:
As the curve cuts the x-axis at (h,0) and (k,0). Therefore h,k are the roots of the quadratic equation.

For the quadratic equation is in the form of ax^2+bx+c=0,

The product of the roots =c/a= 256/1=256 and the sum of the roots = -b/a=-b

256 can be expressed as product of two numbers in the following ways:
1 * 256
2 * 128
4 * 64
8 * 32
16 * 16
The sum of the roots is maximum when the roots are 1 and 256 and the maximum sum is 1 + 256 = 257.

The least value possible for b is therefore -257.

I know this is rather a simple question but i am having trouble understanding the basic concepts. Can someone please elaborate??

Hi,

let me try to explain it to you..

y = x^2 + bx + 256..
when we say the curve cuts the x-axis at (h,0) and (k,0), it means h and k are the roots of equation since here y is 0
x^2 + bx + 256 = 0...
We can see both (h,0) and (k,0) have y as 0, which converts the equation y = x^2 + bx + 256.. as x^2 + bx + 256 = 0...

for a quadratic equation, ax^2+bx+c=0, the product of its roots are c/a and sum= -b/a..
here a=1, b=b and c=256
so here product= hk=c=256..
sum= -b =h+k
sice we are looking for the max sum, so as to have least b, we take h and k as 256 and 1..
P=256*1=256
so SUM = -b= 256+1=257
so b=-257

Thank you so much. This is extremely helpful
Manager
Joined: 09 Jul 2013
Posts: 109
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

17 Mar 2016, 06:45
chetan2u wrote:
davedekoos wrote:
If we were to factor $$x^2+bx+256 =0$$ we would get

$$(x-h)(x-k)=0$$, where h and k are the roots of the equation (as given - "...cuts the x axis at (h, 0) and (k, 0)").

So (-h)*(-k) = 256 and -h-k = b

Possible combinations for h and k are the factor pairs of 256:
1, 256
2, 128
4, 64
etc...

And also the negative of these pairs.
-1, -256
-2, -128
etc.

We can see that to have the smallest value of h+k, they must be -1 and -256

Therefore the lowest value of b = -1-256 = -257

Hi,
you have gone wrong in coloured portion..
For lowest value we require MAX sum as SUM= -b/a
so h and k will be 1 and 256 and not -1 and -256

Yes thank you for pointing that out!

I should have written that b=-(h+k), and therefore to minimize b we must maximize (h+k), and therefore the the values must be 1 and 256. I knew what I was doing, I just wrote it out wrong.

Cheers
_________________

Dave de Koos

Intern
Joined: 23 Sep 2017
Posts: 19
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

12 Nov 2017, 12:55
To the line cut x axis at 2 points, the equation: x^2 + bx +256 = 0 have 2 roots:
From the stimulus, 2 roots are integer => the sum of roots = -b/a = -b integer and the product of roots = c/a = 256 ( theory of equation of degree 2)
It means : h+k = -b; hk = 256
=> the least value of b are equivalent to h+k is the least, or -1 and -256
=> b= -257
Intern
Joined: 28 Dec 2010
Posts: 49
y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

13 Nov 2017, 04:18
1
Simplest Solution.

Since 2 roots exist, $$b^2 - 4ac >0$$
$$b^2 - 32^2>0$$
b>32 & b<-32

Least value among the options is -257.
At this value h and k would be 1 & 256. (both integers)

E.
_________________

_________________________________________

Intern
Joined: 26 Oct 2017
Posts: 27
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar  [#permalink]

### Show Tags

21 Nov 2017, 10:11
Abhishiek wrote:
Simplest Solution.

Since 2 roots exist, $$b^2 - 4ac >0$$
$$b^2 - 32^2>0$$
b>32 & b<-32

Least value among the options is -257.
At this value h and k would be 1 & 256. (both integers)

E.

That's the best solution. Kudos to you.

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar &nbs [#permalink] 21 Nov 2017, 10:11
Display posts from previous: Sort by