false
Catalog
Grades 11-12 Video Solutions 2013
Levels 11&12 Video Solutions 2013 problem24
Levels 11&12 Video Solutions 2013 problem24
Back to course
[Please upgrade your browser to play this video content]
Video Transcription
Question number 24, how many solutions x comma y where x and y are real numbers does the equation x squared plus y squared equal to the absolute value of x plus the absolute value of y have? Said in another way, the question is asking how many points with coordinates x comma y lie on this graph that is described by the equation? So, since we are thinking in that context, let's think of the easiest piece of that graph that lies in quadrant one because then we get to restrict x to be non-negative and y to be non-negative. So, let's look at the graph of this function. Only in quadrant one, so where x is non-negative and y is non-negative. Then we are actually looking at the same equation but with the absolute values removed. And we can move everything to one side, group the x terms with the y terms and complete the square which will give us x squared minus x plus one-fourth plus y squared minus y plus one-fourth is equal to one-half by adding one-fourth twice to both sides. And then the point of doing that was that the expression in parentheses becomes a perfect square. So, we have x minus one-half, y minus one-half quantity squared is equal to one-half. And we recognize this as a circle with center at the point one-half comma one-half and radius are equal to the square root of one-half. So, in quadrant one, this represents an arc of that circle. And so, as a circle contains infinitely many points, the graph of just any arc belonging to that circle will also contain infinitely many points. And so, that's how we interpreted the problem and we conclude that there are infinitely many solutions to our original equation.
Video Summary
The given equation \(x^2 + y^2 = |x| + |y|\) can be analyzed by considering different quadrants. By examining only the first quadrant, where \(x\) and \(y\) are non-negative, the equation simplifies to \(x^2 + y^2 = x + y\). Completing the square transforms it into \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}\), representing a circle with center \((\frac{1}{2}, \frac{1}{2})\) and radius \(\sqrt{\frac{1}{2}}\). Thus, the equation has infinitely many solutions, corresponding to points on the circle.
Keywords
equation
quadrants
circle
solutions
radius
×
Please select your language
1
English