mohan - You will be able to do it BUT you need to think quick.
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?
You needn't always know the formula/e to solve the problem. I will try and solve the problem with out using any formulae.
Since (h,0) and (k,0) lie on the curve y = x^2 + bx + 256, h and k are roots of the equation. So, y can be written in the form (x-h)*(x-k).
y = (x-h)*(x-k) = x^2 + (-h-k)*x + hk = x^2 + bx + 256
Equating the co-efficients of x, we get
(-h-k) = b = -(h+k)
Equating the constants, we get
hk = 256
For b to be minimum, -(h+k) should be minimum.
For (-h-k) to be minimum, h+k should be maximum. We need maximum value h+k where hk = 256
hk = 1*256, h + k = 1 + 256 = 257
hk = 2*128, h + k = 2 + 128 = 130
hk = 4*64, h + k = 4 + 64 = 68
hk = 8*32, h + k = 8 + 32 = 40
hk = 16*16, h + k = 16 + 16 = 32
Never forget to consider the integers on the other side of zero(Though not relevant in this question)
hk = -1*-256, h + k = -1 - 256 = -257
hk = -2*-128, h + k = -2 - 128 = -130
hk = -4*-64, h + k = -4 - 64 = -68
hk = -8*-32, h + k = -8 - 32 = -40
hk = -16*-16, h + k = -16 - 16 = -32.
The highest value of h+k = 257 and the least value of -(h+k) = b = -257.
My GMAT exam scribbling pad would contain just three steps. They are
y = (x-h)*(x-k) = x^2 + bx + 256
b = -(h+k);hk = 256
h = 256 k = 1 -(h+k) = -257
