Sum of squares

Algebra Level 2

Consider $(x + 2y - 8)^2 + (x - y + 1)^2 = 0$ where $x$ and $y$ are real numbers.

What must be the value of x?

2 3 Not enough information

1 solution

Md Omur Faruque
Sep 2, 2015

As squares of real numbers are non-negative, for the sum of $2$ squares to be $0$ , they should both be $0$ . So, we get, $x+2y-8=0$ $x-y+1=0$

Solving these $2$ equations we get, $(x, y) \equiv (\color{#69047E} {\boxed {2}},3)$

Thank you for the solution!

Bharath Murali - 5 years, 9 months ago

I can see you're the author of this question. So, you knew the solution already. Any specific reason for the thanks?

MD Omur Faruque - 5 years, 9 months ago

He's being friendly!

Calvin Lin Staff - 5 years, 9 months ago

@Calvin Lin – Reports, mails, comments, checking problems, giving feedbacks to solutions, slack. I wonder sometime, how do you manage to do all these everyday?

MD Omur Faruque - 5 years, 9 months ago

@Md Omur Faruque – Sleep is for the weak!

Calvin Lin Staff - 5 years, 9 months ago

Oh it's because I forgot to add the solution myself and you did that or me.

Bharath Murali - 5 years, 9 months ago

@Bharath Murali – Oh no! It's my pleasure. Just because you've posted the problem it's not mandatory for you to post a solution too.

MD Omur Faruque - 5 years, 9 months ago

Sum of squares

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