4. Two resistors R1 and R2, when connected in series; their equivalent is 20 Ω , and when connected in parallel; their equivalent is 4.8 ?

a+b=20

1/a+1/b=1/4.8ab=20*4.8=96

a       b       ab            a+b

10    10     1009     96/9=9.667         19.6678  96/8=    12                20

a=8       b=12a=12     b=8

good problem

WolframAlpha gives the same answer.

R1+R2 = 20 ohm

R2 = (20-R1)

4.8 = R1*R2/(R1+R2)

4.8*20 = R1*(20-R1) 

96-20R1+R1^2 = 0

R1 = (20+√ 20^2-96*4)/2 = 12.0 ohm

R2 = (20-√ 20^2-96*4)/2 = 8.0 ohm

check

R1+R2 = 12+8 = 20 ...OK

R1 // R2 = R1*R2 / (R1+R2) = (12*8)/20 = 96/20 = 4.8 ....OK 

R1R2/(R1+R2) = 4.8 but (R1+R2) = 20

R1R2/20 = 4.8 so R1R2 = 96 

Two equations and two unknowns

R2 = 20 - R1

Substitute into R1R2 = 96

R1(20-R1) = 96

-R1²+20R1 - 96 = 0  multiply by -1

R1² -20R1 + 96 = 0

(R1-8)(R1-12) = 0 so R1 = 8 and R2 = 12

Check: 8*12/(8+12) = 96/20 = 4.8Ω Checks

x + y = 20 Ω

xy / (x+y) = 4.8

x = 20 – y

xy / (x+y) = 4.8

(20 – y)y / (20) = 4.8

20y – y² = 96

y² – 20y + 96 = 0

y = 12 or 8

x = 12 or 8

ans, 8 and 12 Ω

x + y = 20

1/x + 1/y = 1/4.8

Solve for x and y.