Question

The product of two numbers added to their sum.​

Answer 1

Answer:

If you are asked to work out the product of two or more numbers, then you need to multiply the numbers together. If you are asked to find the sum of two or more numbers, then you need to add the numbers together.

Answer 2

Sorry, your question is not understable.

Here's an example for your question.

Question;

The product of two numbers is 12 .. If their sum added to the sum of their squares is 32, find the numbers ?

Answer;

Product of numbers = 12

Possibilities (1,12) , (2,6), (3,4)

Sum of number + sum of squares = 32

Hence 1st and second combinations aren't possible.

So the numbers are 3,4

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Let the numbers be a & b

Given,

$a \: \times \: b \: = \: 12$

$a + b + ( {a}^{2} + {b}^{2} ) \: = \: 32$

... ... ... ... .. ... ... ... ... ... ... ... ... ... .. ..

(a^2+b^2) = (a+b)^2 - 2*a*b [identity]

$( {a}^{2} \: + \: {b}^{2} ) \: = \: (a \: + \: b) \: - \: {(a + b)}^{2} - 2 \times a \times b \: \: \: \: |identity|$

$(a+b) + (a+b)^2 - 2×a×b = 32$

$(a+b) + (a+b)^2 - 2×12 = 32$

$(a+b) + (a+b)^2 = 56$

$(a+b)^2 + (a+b) - 56 = 0 \\ </p><p>(a+b)^2 + 8(a+b) - 7(a+b) - 56 = 0 \\ </p><p>(a+b) × [(a+b) + 8] - 7 × [(a+b) + 8] = 0$

$[(a+b) + 8] × [(a+b)-7] = 0$

$a+b = 7$

$ab = 12$

$b = 12/a$

$a + 12/a = 7$

$a^2 + 12 = 7a$

$a^2 - 7a + 12 = 0$

$(a-4)×(a-3) = 0$

$a=4 \: \: \: \: or \: \: \: 3$

$So \: \: \: if \: \: \: a = 3, b=4$

$Hence \: \: \: numbers \: \: \: are \: \: \: 3,4$