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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Answered by
60
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)
Answered by
132
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.
thank you
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
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