Find Intersection of -sqrt(x) & x-6 Functions

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Homework Help Overview

The problem involves finding the intersection points of the functions y = -sqrt(x) and y = x - 6. Participants are exploring methods to solve the equation formed by setting these two functions equal to each other.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss equating the functions and manipulating the resulting equation. There are attempts to simplify the equation by squaring both sides and factoring. Some participants question the validity of potential solutions, particularly regarding extraneous solutions introduced by squaring.

Discussion Status

The discussion is active, with various approaches being explored, including direct manipulation of the equation and substitution methods. Participants are engaging with the implications of their algebraic manipulations, particularly concerning the verification of solutions against the original equations.

Contextual Notes

There is an emphasis on checking solutions for extraneous results, particularly after squaring both sides of the equation. Participants are also navigating the implications of their algebraic transformations and the conditions under which certain solutions may not hold true.

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Homework Statement



I need to find the point at which these functions intersect:
y = -sqrt(x)
y = x - 6


Homework Equations





The Attempt at a Solution



I set them to equate:
-sqrt(x) = x - 6
-sqrt(x) - x + 6 = 0

Now how do I find the root?
 
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equate them tgt , [tex]\sqrt{x}=6-x[/tex] .
Now square both sides , x=36+x²-12x ... continue from here.
 
Do I move them all to one side and get x^2 -13x + 36 = 0?
 
Yes. Now factor.

Be sure to check each solution in the original equation, though. When you square both sides of an equation, extraneous solutions are sometimes introduced, values that are not solutions of your original equation.
 
Alternatively, you can say u=sqrt(x) and rewrite the equation as -u-u^2+6=0
 
Okay nice.
So I can't use x = 9 because -sqrt(9) not= 9-6, correct?
 
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
 
Thanks guys!
 
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
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.
 

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