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Bunuel
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   09 Mar 2016, 12:57
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E
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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
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E
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   10 Mar 2016, 14:56
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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

Answer: E
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   10 Mar 2016, 05:15
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Bunuel
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
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   10 Mar 2016, 22:53
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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.

Answer " E
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   17 Mar 2016, 04:34
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AshokGmat2016
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.

Answer " E

I know this is rather a simple question but i am having trouble understanding the basic concepts. Can someone please elaborate??
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   17 Mar 2016, 05:17
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AshokGmat2016
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.

Answer " E

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
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   17 Mar 2016, 05:20
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davedekoos
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

Answer: E

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
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   17 Mar 2016, 05:27
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AshokGmat2016
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.

Answer " E

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 :)
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   17 Mar 2016, 06:45
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davedekoos
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

Answer: E

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
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   12 Nov 2017, 12:55
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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
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   13 Nov 2017, 04:18
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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.
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   21 Nov 2017, 10:11
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Abhishiek
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
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   03 Nov 2019, 04:34
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Can someone please help?

Since the equation has two distinct real roots, why can't I solve the question by using the formula:
b^2-4ac>0?
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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar   04 Jul 2023, 14:11
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The quadratic equation x^2+ bx+256 has two solutions, therefore the discriminant b^2 - 4ac must be positive.

b^2 - 4ac>0
b^2 - (4x1x256)>0
b^2 - 1024>0
b^2>1024
sqrt b^2> sqrt 1024
∣b∣ > sqrt 1024
∣b∣ > 32
b>32 or b<-32

The question asks what is the least value of b considering that h and k are integers.

We can discard 0 and -32 as they are outside the range of the possible values of b.

Also (applying Vieta's theorem):
h x k=256
and -(h+k)=b

the only 2 integers that will multiply to 256 and add up to constitute a number that appears in the answer choices is -256 and -1

Therefore, b= -256 - 1= -257.

Correct answer is E
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