If the sum of two numbers is 13 and the product of the same two numbers is 40, find the sum of the squares of the two numbers without finding the actual numbers.
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So:
x+y = 13
x*y = 40
You then know that (a+b)^2 = a^2+2ab+b^2, so ... (a+b)^2 -2ab = a^2+b^2, which is the sum you want ...
(x+y)^2 = 13^2
2*x*y = 80, therefore x^2+y^2 (the sum of the squares of the two numbers) = 13^2-80 = 89
x+y = 13
x*y = 40
You then know that (a+b)^2 = a^2+2ab+b^2, so ... (a+b)^2 -2ab = a^2+b^2, which is the sum you want ...
(x+y)^2 = 13^2
2*x*y = 80, therefore x^2+y^2 (the sum of the squares of the two numbers) = 13^2-80 = 89
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(a + b) = 13
(a + b)² = 13²
a² + b² + 2ab = 139
a² + b² = 139 – 2ab
a² + b² = 139 – 80 = 59
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(a + b)² = 13²
a² + b² + 2ab = 139
a² + b² = 139 – 2ab
a² + b² = 139 – 80 = 59
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x + y = 13 and xy =40
(x + y)^2 = x^2 + 2xy + y^2
13^2 - 2*40 = x^2 + y^2
169 - 80 = x^2 + y^2 = 89
(x + y)^2 = x^2 + 2xy + y^2
13^2 - 2*40 = x^2 + y^2
169 - 80 = x^2 + y^2 = 89
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89
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Let a and b be the numbers.
a^2 + b^2 = (a+b)^2 - 2ab = 13^2 - 2 x 40 = 169 - 80 = 89
a^2 + b^2 = (a+b)^2 - 2ab = 13^2 - 2 x 40 = 169 - 80 = 89
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The answer is 25+64=89