if AM of the roots of a quadratic equation is 8/5and AM of their reciprocal is 8/7
Answers
Answer:
The arithmetic mean of the roots is 8/5.
So we can think of the roots as:
r = 8/5 - d
s = 8/5 + d
Their reciprocals are:
1/r = 1/(8/5 - d)
1/s = 1/(8/5 + d)
The arithmetic mean of those two numbers is:
(1/r + 1/s) * 1/2 = [ (8/5 + d) + (8/5 - d) ] / 2[ (8/5 - d)(8/5 + d) ]
= (16/5) / 2((8/5)² - d²)
(8/5) / ((8/5)² - d²) = 8/7
Cross multiply:
8*((8/5)² - d²) = 7*8/5
(8/5)² - d² = 7/5
d² = (8/5)² - 7/5
d² = 64/25 - 35/25)
d² = 29/25
d = ±√29/5
So we have:
x = (8 ± √29) / 5
2a = 5
a = 5/2
-b = 8
b² - 4ac = 29
64 - 4(5/2)c = 29
64 - 10c = 29
35 = 10c
c = 7/2
Answer:
f(x) = (5/2)x² - 8x + 7/2
or
f(x) = ½(5x² - 16x + 7)
We can also multiply that by 2 to get the same roots:
f(x) = 5x² - 16x + 7
let the roots be x&y
x+y / 2 = 8/5
x+y = 16/5. (1)
(1/x + 1/y) / 2 = 8/7
(x+y)/2xy = 8/7
16/5 = (16xy)/7
xy = 7/5
x = 7/5y
put it in (1)
7/(5y) + y = 16/5
5y²+7 = 16y
5y² - 16y +7 = 0