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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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09 Mar 2016, 12:57
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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
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10 Mar 2016, 14:56
If we were to factor \(x^2+bx+256 =0\) we would get \((xh)(xk)=0\), 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 hk = 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 = 1256 = 257 Answer: E
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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
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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 coordinate 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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Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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10 Mar 2016, 22:53
As the curve cuts the xaxis 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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Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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17 Mar 2016, 04:34
AshokGmat2016 wrote: As the curve cuts the xaxis 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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Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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17 Mar 2016, 05:17
MeghaP wrote: AshokGmat2016 wrote: As the curve cuts the xaxis 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 xaxis 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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Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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17 Mar 2016, 05:20
davedekoos wrote: If we were to factor \(x^2+bx+256 =0\) we would get
\((xh)(xk)=0\), 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 hk = 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 = 1256 = 257
Answer: E Hi, you have gone wrong in coloured portion.. For lowest value we require MAX sum as SUM= b/aso h and k will be 1 and 256 and not 1 and 256
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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
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17 Mar 2016, 05:27
chetan2u wrote: MeghaP wrote: AshokGmat2016 wrote: As the curve cuts the xaxis 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 xaxis 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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Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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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
\((xh)(xk)=0\), 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 hk = 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 = 1256 = 257
Answer: E Hi, you have gone wrong in coloured portion.. For lowest value we require MAX sum as SUM= b/aso h and k will be 1 and 256 and not 1 and 256Yes 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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Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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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



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y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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13 Nov 2017, 04:18
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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Re: y = x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k ar
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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




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