The sum of a number and its reciprocal is 41/20. Find the numbers. smaller value larger value

Step-by-step explanation:

Let the number be x.

The reciprocal of the number will be 1/x

If the sum of the number and its reciprocal is 41/20, this can be represented as;

Uisng the general formula to get x

x = -b±√b²+4ac/2a

x = 41±√41²-4(20)(20)/2(20)

x = 41±√1681-1600/40

x = 41±√81/40

x = 41±9/40

x = 50/40 or 32/40

x = 5/4 or 4/5

if the value is 5/4, the other value will be 4/5

The numbers are 5/4 and 4/5

The smaller value is 4/5

The larger value is 5/4

a=5/4 or 4/5

Therefore, the smaller value = 4/5

The larger value = 5/4

Step-by-step explanation:

Let the number be represented by a

And it's reciprocal be represented by 1/a

So we have

a + 1/a = 41/20

Cross Multiply

20( a +1/a) = 41

20a +20/a =41

Find the LCM which is a

20a² + 20 = 41a

20a² + 20 - 41a =0

20a² - 41a +20 = 0

20a²-25a - 16a + 20 =0

5a(4a - 5) -4( 4a - 5) = 0

(5a - 4)(4a - 5) = 0

5a - 4 = 0

5a = 4

a = 4/5

or

4a - 5 = 0

4a =5

a = 5/4

Therefore, the number which is represented by a is

1) a = 4/5 while it's reciprocal which is 1/a is 5/4

or

2) a = 5/4 which it's reciprocal which is 1/a = 4/5

Therefore, the smaller value = 4/5

The larger value = 5/4

c(x)= 6x + 3

step-by-step explanation: