y = 2x^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? Pick:  A) -120.5  B) -124.5  C) -128.5  D) -132.5  E) 132.5
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= 256/2= 
<gadget id="calculator">256/2</gadget>
<output>128</output>
128 and the sum of the roots=-b/a=-b   128 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 1/2 and 128 and the maximum sum is 1/2 + 128= 
<gadget id="calculator">1/2 + 128</gadget>
<output>257/2 = around 128.5</output>
128.5   The least value possible for b is therefore -128.5.   -128.5
<result>C</result>
y = 2x^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?