Question

The sum of two numbers is 9 and the sum of their squares is 41. Taking one number
as x, form an equation in x and solve it to find the numbers.
mal in thnnn less than twice the square of the​

Answer 1

Answer:

Step-by-step explanation:

The answer is 178

give a thanks

Answer 2

one number be x another be y then x+y=9

x=9-y.........(1)

x^2+y^2=41

..........(2)

substituting .(1) in.(2)

(9-y)^2+y^2=41

9^2-2×9×y+y^2+y^2=41

81-18y+2y^2=41

2y^2-18y+81-41=0

2y^2-18y+40=0÷2

y^2-9y+20=0

a=1 b=-9 c=20

substituting values on quadratic formula

-b+-root b^2-4ac%2a

we get 5 and 4

therefore the numbers are 5 and 4