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

Step-by-step explanation:

Given

$h(t)=(139kt)^2-69t+80$

For roots $h(t)=0$

$(139kt)^2-69t+80=0\\t=\dfrac{69\pm\sqrt{(-69)^2-4\times (139k)^2(80)}}{2\times 139}$

$t=\dfrac{69\pm\sqrt{4761-6182720k^2}}{278}$

Larger root: $t=\dfrac{69+\sqrt{4761-6182720k^2}}{278}$

smaller root: $t=\dfrac{69-\sqrt{4761-6182720k^2}}{278}$

For exactly one root D=0

i.e. $4761-6182720k^2=0\\\\k=\dfrac{69}{139\times 4\times 2\times \sqrt{5}}\\k=0.0277$