Question

Find two numbers whose sum is 4 and it's product is 8.

Answer 1

Answer:

xy = 4 => y = 4/x => 1/y = x/4

1/x + 1/y = 65/56

1/x + x/4 = 65/56

Multiply both sides by 56x:

56 + 14x^2 = 65x

14x^2 - 65x + 56 = 0

[I chose to find Vertex Form and solve that.]

h = -b/2a = - -65/(2*14) = 65/28

k = c - ah^2 = 56 - 14(65/28)^2

= (56^2 - 65^2)/56 = (56-65)(56+65)/56 = -9(121)/56

x = h ± √(-k/a) = 65/28 ± √(- -9(121)/(56*14))

= 65/28 ± √(3^2(11^2)/(4*14^2))

= 65/28 ± 3*11/(2*14) = (65 ± 33)/28

x = 7/2 or 8/7

y = 4/x = 8/7 or 7/2

So the two numbers are 8/7 and 7/2.

check:

xy =? 4

(8/7)(7/2) =? 4

4 = 4 √

1/x + 1/y =? 65/56

1/(8/7) + 1/(7/2) =? 65/56

7/8 + 2/7 =? 65/56

7*7/(8*7) + 2*8/(7*8) =? 65/56

(49 + 16)/56 =? 65/56

65/56 = 65/56 √

Answer 2

Answer:

so the two number is 2and 4=2+4=6

2×4=8