Question

Find two numbers (exactly) whose product is 10 and whose sum is 18.

Answer 1

a = first number

b = second number


Conditions :

a * b = 10

a + b = 18



a + b = 18 Subtract a to both sides

a + b - a = 18 - a

b = 18 - a


a * b = 10

a * ( 18 - a ) = 10

18 a - a ^ 2 = 10

- a ^ 2 + 18 a = 10 Multiply both sides by - 1

a ^ 2 - 18 a = - 10 [ Add ( 18 /2 ) ^ 2 = 9 ^ 2 = 81 ] to both sides

a ^ 2 - 18 a + 81 = - 10 + 81

a ^ 2 - 18 a + 81 = 71

( a - 9 ) ^ 2 = 71

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Becouse :

( a - 9 ) ^ 2 = a ^ 2 - 2 a * 9 + 9 ^ 2 = a ^ 2 - 18 + 81
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( a - 9 ) ^ 2 = 71 Take square root to both sides

a - 9 = + OR - sqroot ( 71 ) Add 9 to both sides

a - 9 + 9 = 9 + OR - sqroot ( 71 )

a = 9 + OR - sqroot ( 71 )


The solutions are :

a = 9 - sqroot ( 71 )

and

a = 9 + sqroot ( 71 )


Now you have two set of solutions of this problem :

1 )

a = 9 - sqroot ( 71 )

b = 18 - a

b = 18 - [ 9 - sqroot ( 71 ) ]

b = 18 - 9 + sqroot ( 71 )

b = 9 + sqroot ( 71 )


2 )

a = 9 + sqroot ( 71 )

b = 18 - a

b = 18 - [ 9 + sqroot ( 71 ) ]

b = 18 - 9 - sqroot ( 71 )

b = 9 - sqroot ( 71 )


Final solutions :

1 )

a = 9 - sqroot ( 71 )

b = 9 + sqroot ( 71 )


2 )

a = 9 + sqroot ( 71 )

b = 9 - sqroot ( 71 )

Answer 2

You want to find two numbers whose product is 10 and whose sum is 18.

So solution 1 and solution 2 are same solution.


The numbers are:


9 - sqroot ( 71 )

and

9 + sqroot ( 71 )

Answer 3

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