But x = 9 is not a solution of sqrt(x)= 6 -x, which is what you started with. x = 9 is an extraneous solution that I warned you of.zeion said:Ah okay, that u substitution seems easier:
Let u = sqrt(x), then
u^2 + u - 6 = 0
(u+3)(u-2) = 0
u = -3, u = 2
sqrt(x) = -3, sqrt(x) = 2
x = 9, x = 4
The "Find Intersection of -sqrt(x) & x-6 Functions" problem asks for the values of x where the two functions, f(x) = -sqrt(x) and g(x) = x-6, intersect. In other words, we need to find the value(s) of x that make both functions equal to each other.
To solve this problem, we need to set the two functions equal to each other and then solve for x. This can be done using algebraic methods, such as isolating x on one side of the equation and simplifying. Once we have the value(s) of x, we can plug them back into the original functions to find the corresponding y-values.
Yes, there are restrictions when solving this problem. Since the function f(x) = -sqrt(x) is undefined for negative values of x, we can only find intersections for values of x that make both functions defined. In this case, x must be greater than or equal to 0.
Yes, we can use a graphing calculator or software to graph the two functions and visually find the points where they intersect. This can give us an approximate solution, but it is important to note that it may not be exact due to the limitations of graphing technology.
Yes, it is possible for this problem to have more than one solution. Since the two functions are not parallel, they can intersect at multiple points. It is important to check all potential solutions to ensure that they are valid and satisfy the restrictions mentioned in question 3.