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The difference of two numbers is 1. What is the smallest possible value for the sum of their squares?​

Respuesta :

fuller14
X= 1 number
Y= a number that is 1 less than x
Y= x-1

The sum of squares=

X^2 + (x-1)^2 = x^2 + x^2 - 2x + 1
X^2 + x^2 - 2x + 1 = 2x^2 - 2x + 1

Assuming that integers = numbers. We could try some values

When x = 0, the sum of squares = 1
When x = -1, the sum of squares = 4
When x = 1, the sum of squares = 1

If you put this into a graph, it is a parabola and we see that the minimum value hits at x = 1/2.

To check this value:

2x^2 - 2x + 1 = 2·(1/4) - 2·(1/2) + 1
2·(1/4) - 2·(1/2) + 1 = 1/2 - 1 + 1
1/2 - 1 + 1 = 1/2

When you look back at the question, it asks for 2 answers

X= 1/2
X-1 = -1/2

What is the sum of their squares?

1/4 + 1/4 = 1/2

That fits with graph too.

Answers: 1/2 and -1/2

Hope this helps!!

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