Explanation:
Let [tex]R_1[/tex] and [tex]R_2[/tex] be the the resistances of the resistors. We are given that
[tex]R_1 + R_2 = 690\:Ω\:\:\:\:\:\:\:(1)[/tex]
and
[tex]\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{118\:Ω}\:\:\:\:\:(2)[/tex]
From Eqn(1), we can write
[tex]R_2 = 690\:Ω - R_1[/tex]
and then plug this into Eqn(2):
[tex]\dfrac{1}{R_1} + \dfrac{1}{690\:Ω - R_1} = \dfrac{1}{118\:Ω}[/tex]
or
[tex]\dfrac{690\:Ω}{(690\:Ω)R_1 - R_1^2}= \dfrac{1}{118\:Ω}[/tex]
[tex]\Rightarrow R_1^2 - (690\:Ω)R_1 + (690\:Ω)(118\:Ω)= 0[/tex]
or
[tex]R_1^2 - 690R_1 + 81420 = 0[/tex]
Using the quadratic formula, we find that the above equation has two roots:
[tex]R_1 = 151.1\:Ω,\:\:538.9\:Ω[/tex]
This means that if you choose one root value for [tex]R_1[/tex], the other root will be the value for [tex]R_2[/tex].
