The sum of two numbers is18 and the sum of their squares is 164. Find the numbers.

Answer 1

Lets the numbers are x and y

so see the conditions above , sum of 2 no s is 8 , so we can write x + y = 18...........eqn . no (1)

another condition above, sum of the square of numbers 164 so x^2+ y^2 = 164....eqn no (2)

from eqn no 1. x + y = 18

subtracting y from both side,

x = 18-y now substituting this value of x in eqn. no (2) we get,

(18- y )^2 + y^2 = 164..

expanding the whole square formula ,

(18)^2- 2*18*y + y^2 + y^2 = 164

324 - 36y + 2y^2 = 164

subtracting 164 from both sides,

324 - 36y + 2y^2 - 164 = 164 - 164

2y^2 - 36y + 160 = 0

dividing by 2 we get , y^2 - 18y + 80= 0

y^2 -10y - 8y + 80=0

y( y-10) -8( y-10)= 0

( y-10 ) (y-8 ) = 0

so either y = 10 or y= 8

two no.s are 10 and 8

Answer 2

You could make 2 equations, x+y=18 and x^2+y^2 = 164 and solve that system of equations.

Or you just try a couple numbers, integers, 9+9=18, but 81+81=162. that's close. Try 8 and 10 = 18

and 64+100 = 164.

1+17=18 and their squares total 289. the end extremes don't work, so try something in the middle and you stumble on 8 and 10 as the answer.

The only possibilities are

0 and 18

1 and 17

2 and 16

3 and 15

4 and 14

5 and 13

6 and 12

7 and 11

8 and 10

9 and 9

10 possible solutions, try one or two and you get an idea how close you are.

Answer 3

let's say we have two numbers, x and y

their sum is 18 therefore x+y=18

the sum of their squares is 164 therefore x²+y²=164

let's solve for x in terms of y from the first equation and plug that into our second equation...

x+y=18

x=18-y

(18-y)²+y²=164

18²-36y+y²+y²=164

2y²-36y+324=164

y²-18y+162=82

y²-18y+80=0

(y-8)(y-10)=0

y=8 or y=10

plugging that into our first equation tells us x = 8 when y = 10 or x = 10 when y = 8

our two numbers are 8 and 10