Question

Find the roots of h(t) = (139kt)^2 − 69t + 80

the smaller root is:
the larger root is:

The answers will consist of algebraic expressions containing the parameter k.

What positive value of k will result in exactly one real root?
K = ?

Answer 1

The positive value of
`k` will result in exactly one real root is approximately 0.028.

Explanation:

Let
`h(t) = 19321\cdot k^(2)\cdot t^(2)-69\cdot t +80`, roots are those values of
`t` so that
`h(t) = 0`. That is:

`19321\cdot k^(2)\cdot t^(2)-69\cdot t + 80=0` (1)

Roots are determined analytically by the Quadratic Formula:

`t = \frac{69\pm \sqrt{4761-6182720\cdot k^(2) }}{38642}`

`t = (69)/(38642) \pm \sqrt{(4761)/(1493204164)-(80\cdot k^(2))/(19321) }`

The smaller root is
`t = (69)/(38642) - \sqrt{(4761)/(1493204164)-(80\cdot k^(2))/(19321) }`, and the larger root is
`t = (69)/(38642) + \sqrt{(4761)/(1493204164)-(80\cdot k^(2))/(19321) }`.

`h(t) = 19321\cdot k^(2)\cdot t^(2)-69\cdot t +80` has one real root when
`(4761)/(1493204164)-(80\cdot k^(2))/(19321) = 0`. Then, we solve the discriminant for
`k`:

`(80\cdot k^(2))/(19321) = (4761)/(1493204164)`

`k \approx \pm 0.028`

The positive value of
`k` will result in exactly one real root is approximately 0.028.