Lesson Solving one tricky non-linear equation
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<H2>Solving one tricky non-linear equation</H2> <H3>Problem 1</H3>Solve an equation floor(x^2 + 2x + 5) - floor(x + 1) = 6, where the symbol 'floor' denotes the greatest integer function (floor function). <B>Solution</B> <pre> For clarity, I will write this equation in this form floor(x^2 + 2x + 5) - floor(x+1) = 6. (1) I will solve it step by step. (1) x^2 + 2x + 5 = (x+1)^2 + 4. Therefore, floor(x^2 + 2x + 5) = floor((x+1)^2 + 4). Next, it is clear that floor((x+1)^2+4) = floor((x+1)^2) + 4. Therefore, equation (1) is equivalent to floor((x+1)^2) + 4 - floor(x+1) = 6, or floor((x+1)^2 - floor(x+1) = 2. (2) Now I will introduce new variable u = x+1 to simplify writing. So, now an equation to solve is floor(u^2) - floor(u) = 2. (2) (3) For what will follow, it is convenient to look at the plot of the functions u --> y = floor(u^2) - floor(u) and y = 2. These functions are shown in the plot https://www.desmos.com/calculator/9eg9wmaonb (4) In the interval -1 < u < 0, floor(u^2) = 0 and floor(u) = -1, so floor(u^2) - floor(u) = 0 - (-1) = 1, which means that there are no solutions to equation (2) in this interval. (5) At point u = -1, floor(u^2) = 1; floor(u) = -1, so floor(u^2) - floor(u) = 1 - (-1) = 2, which means that u = -1 is a solution to equation (2). (5) In the interval -2 < u < -1, floor(u^2) has values 1 or 2 or 3; floor(u) = -2, so floor(u^2) - floor(u) has values 3, or 4, or 5, which means that there are no solutions to equation (2) in this interval. At this point, it is clear that there is no sense to analyze further for negative 'u' lesser than -2, because there are no solutions to equation (2) there. (6) In the interval 0 < u < 1, floor(u^2) has value 0; floor(u) has value 0, so floor(u^2) - floor(u) = 0 - 0 = 0, which means that there are no solutions to equation (2) in this interval. (7) At point u = 1, floor(u^2) = 1; floor(u) = 1, so floor(u^2) - floor(u) = 1 - 1 = 0, which means that u = 1 is not a solution to equation (2). (8) In the interval 1 < u < {{{sqrt(2)}}}, floor(u^2) has value1 1; floor(u) = 1, so floor(u^2) - floor(u) = 1 - 1 = 0, which means that there are no solutions to equation (2) in this interval. (9) At point u = {{{sqrt(2)}}}, floor(u^2) = 2; floor(u) = 1, so floor(u^2) - floor(u) = 2 - 1 = 1, which means that u = {{{sqrt(2)}}} is not a solution to equation (2). (10) In the interval {{{sqrt(2)}}} < u < {{{sqrt(3)}}}, floor(u^2) has value 2; floor(u) = 1, so floor(u^2) - floor(u) = 2 - 1 = 1, which means that there are no solutions to equation (2) in this interval. (11) At point u = {{{sqrt(3)}}}, floor(u^2) = 3; floor(u) = 1, so floor(u^2) - floor(u) = 3 - 1 = 2, which means that u = {{{sqrt(3)}}} is a solution to equation (2). (12) In the interval {{{sqrt(3)}}} <= u < 2, floor(u^2) has value 3; floor(u) = 1, so floor(u^2) - floor(u) = 3 - 1 = 2, which means that all this interval is the solution to equation (2). (13) At point u = 2, floor(u^2) = 4; floor(u) = 2, so floor(u^2) - floor(u) = 4 - 2 = 2, which means that u = 2 is a solution to equation (2). (14) In the interval 2 < u < {{{sqrt(5)}}}, floor(u^2) has value 4; floor(u) = 2, so floor(u^2) - floor(u) = 4 - 2 = 2, which means that all this interval is the solution to equation (2). (15) At point u = {{{sqrt(5)}}}, floor(u^2) = 5; floor(u) = 2, so floor(u^2) - floor(u) = 5 - 2 = 3, which means that u = {{{sqrt(5)}}} is not a solution to equation (2). At this point, it is clear that there is no sense to analyze further for positive 'u' greater than {{{sqrt(5)}}}, because there are no solutions to equation (2) there. At this point, we can summarize the solution set for variable u: it is the union of sets of real numbers < -1 > U [ {{{sqrt(3)}}} <= u < {{{sqrt(5)}}} ). Now we can return to variable x and to describe the solution set for the given equation (1) in terms of x. It is < -2 > U [ {{{sqrt(3)-1}}} <= x < {{{sqrt(5)-1}}} ). <<<---=== <U>ANSWER</U> </pre> Thus the problem is solved completely. ///////////////////////////////////////////////////// What to do if in the future you will get a similar problem ? First, make a plot: it will say you a lot about the possible solution. Then start analyze, and pay a special attention analyzing all possible relevant intervals and all possible points of discontinuity of the floor function at every term in the equation. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/OVERVIEW-of-lessons-on-solving-non-linear-equations.lesson>OVERVIEW of lessons on solving non-linear equations in one unknown</A> to navigate over the lessons to this topic. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-II.
