# 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.

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The equation is x²-9x+20=0

Now the value of x is either 5 or 4

so the numbers are (5,4) and (4,5)

Now the value of x is either 5 or 4

so the numbers are (5,4) and (4,5)

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.

thank you