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

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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

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

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10 Mar 2016, 22:53

1

This post received KUDOS

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.

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

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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.

Answer " E

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

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
_________________

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 _________________

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

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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.

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

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

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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

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.

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

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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