SOLUTION: {{{ y = 6x^2 + bx + 11 }}} What is the smallest positive integer for b that will force this parabola to have two x-intercepts? Please show working

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: {{{ y = 6x^2 + bx + 11 }}} What is the smallest positive integer for b that will force this parabola to have two x-intercepts? Please show working      Log On


   

Question 985667: +y+=+6x%5E2+%2B+bx+%2B+11+
What is the smallest positive integer for b that will force this parabola to have two x-intercepts?
Please show working
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
That will be when the discriminant is positive:

bē-4ac = bē-4(6)(11) > 0
              bē-264 > 0
                  bē > 264
              
Taking positive square roots:

                   b > 16.24807681

Smallest positive integer is the next higher positive integer, b=17.

When b=16, no x-intercepts:

graph%28200%2C200%2C-3%2C3%2C-2%2C4%2C6x%5E2%2B16x%2B11%29

When b=16.24807681, 1 x-intercept:

graph%28200%2C200%2C-3%2C3%2C-2%2C4%2C6x%5E2%2Bsqrt%28264%29x%2B11%29

When b=17, 2 x-intercepts:

graph%28200%2C200%2C-3%2C3%2C-2%2C4%2C6x%5E2%2B17x%2B11%29

Edwin