Sum of two numbers is 40.sum of their reciprocals is 2/5.Find the numbers
Answers
let the no. be a and b
a+b=40
b=40-a....1
1/a+1/b=2/5 or
5b+5a=2ab or
b(2a-5)=5a or
b=5à/(2a-5)....2
equate 1 and 2
40-a=5a/(2a-5)
(40-a)(2a-5)=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=2.6795192
Answer:
x^2-40x+100=0
Step-by-step explanation:
let one of the two numbers be X.
then, the other number= 40-X (:. the sum of two numbers is 40)
their reciprocal are 1/X and 1/40-X
therefore, the sum of their reciprocals
= 1/X + 1/40-X
according to the problem situation ,
1/X + 1/40-X = 2/5
taking Lcm:-
40-x+x/X(40-x)=2/5
40/X(40-x)=2/5
dividing both sides by 2,
20/X(40-x)=1/5
Cross multiplying
X(40-x)=(20) (5)
40x-x^2=100
-x^2+40x-100=0
multiple by -1,x^2-40x+100=0
therefore, one of the two numbers satisfies the quadratic equations
x^2-40x+100=0