Question

under the QUADRATIC EQUATION(Roots)
the sum of the reciprocals of two numbers is 3/8 . Find the other number , if one is 4 greater than the other.

Answer 1

we can cheat if you like and just reason it out. since
\[\frac{3}{8}=\frac{2}{8}+\frac{1}{8}=\frac{1}{4}+\frac{1}{8}\] one number is 4, the other is 8

Answer 2

otherwise you have to do a bunch of work. namely put
\[\frac{1}{x}+\frac{1}{x+4}=\frac{3}{8}\] add on the left to get
\[\frac{x+4+x}{x(x+4)}=\frac{2x+4}{x(x+4)}=\frac{3}{8}\] the cross multiply to get
\[3x(x+4)=8(2x+4)\] multiply out and get
\[3x^2+12x=16x+32\] set equal to zero as
\[3x^2-4x-32=0\] factor to get
\[(x-4)(3x+8)=0\] and finally solve for x to get
\[x=4,x=-\frac{8}{3}\] what a pain. now lets check the answers

Answer 3

thank you:))

Answer 4

careful because actually my "think method" did not give me both answers. there are two answers namely
\[x=4\] or
\[x=-\frac{8}{3}\] both work

Answer 5

ok:)