x+y=40
(1/x)+(1/y)=(2/5)
=>x+y=(2/5)xy
=>40(5/2)=40y-y^2[ since x+y=40 =>xy+y^2=40y]
=>y=20+5(12)^(1/2)=a or 20-5(12)^(1/2)=b[pythagorus theorem]
now if x=a then y=b & vice-versa
The sum of two numbers is 40.we need to find the no. ,If the sum of their reciprocal is 2/5
( Quadratic equation)
7 answers
Not understanding the way explained.
Not so good 👎
Let the numbers be a and b.
a+b = 40, or
b = 40-a …(1)
(1/a)+(1/b) = (2/5), or
5b+5a = 2ab, or
b(2a-5) = 5a or
b = 5a/(2a-5) …(2)
Equate (1) and (2)
40-a = 5a/(2a-5)
(40-a)(2a-5) = 5a
80a-2a^2–200+5a = 5a
2a^2–80a+200 = 0
a^2–40a+100 = 0
a1 = [40 +(1600–400)^0.5]/2
= [40+34.64101615]/2
= 37.32050808
b1 = 40-37.32050808 = 2.67949192
a+b = 40, or
b = 40-a …(1)
(1/a)+(1/b) = (2/5), or
5b+5a = 2ab, or
b(2a-5) = 5a or
b = 5a/(2a-5) …(2)
Equate (1) and (2)
40-a = 5a/(2a-5)
(40-a)(2a-5) = 5a
80a-2a^2–200+5a = 5a
2a^2–80a+200 = 0
a^2–40a+100 = 0
a1 = [40 +(1600–400)^0.5]/2
= [40+34.64101615]/2
= 37.32050808
b1 = 40-37.32050808 = 2.67949192
I can't understand can I look along with the video
I can't understand can I see along the video
I think so🤔🤔.... The answer is wrong
And also I can't understand the way you written......
And also I can't understand the way you written......