163k views
21 votes
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 = ?

User TallChuck
by
3.6k points

1 Answer

14 votes

Explanation:

Given


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

For roots
h(t)=0


(139kt)^2-69t+80=0\\t=(69\pm√((-69)^2-4* (139k)^2(80)))/(2* 139)\\


t=(69\pm√(4761-6182720k^2))/(278)

Larger root:
t=(69+√(4761-6182720k^2))/(278)

smaller root:
t=(69-√(4761-6182720k^2))/(278)

For exactly one root D=0

i.e.
4761-6182720k^2=0\\\\k=(69)/(139* 4* 2* √(5))\\k=0.0277

User Kieran Benton
by
3.4k points