SOLUTION: {{{ x^2- sqrt( 11x )- 1 = 0}}}

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Question 714430: +x%5E2-+sqrt%28+11x+%29-+1+=+0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E2-+sqrt%28+11x+%29-+1+=+0
First a bit of vocabulary. "Radicand" is the name for the expression inside a radical.

Here's a procedure for solving equations where the variable is in the radicand of a square root:
  1. Isolate a square root that has the variable in its radicand.
  2. Square both sides of the equation. Squaring the isolated square root will be easy. Squaring the other side can easily be done incorrectly! This is where many mistakes occur so be careful!
  3. If there are still any square roots with the variable in its radicand, then repeat steps 1-3.
  4. At this point there should no longer be any square roots with the variable in its radicand. Use appropriate techniques for whatever kind of the equation the equation now is.
  5. Check your solution(s). This is not optional! Whenever you square both sides of an equation, like step 2, "extraneous" solutions may occur. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. Extraneous solutions may occur any time both sides of an equation is raised to an even power (like squaring). They are not an indication of an error. You just have to check for them and, if found, reject them.
Let's see this in action:
1. Isolate a square root.
There's only one square root so there's not much choice. Adding the square root to each side gives us:
+x%5E2-+1+=++sqrt%28+11x+%29

2. Square both sides.
+%28x%5E2-+1%29%5E2+=++%28sqrt%28+11x+%29%29%5E2
Squaring the right side is simple. Squaring the left side correctly requires either FOIL or use of the %28a-b%29%5E2+=+a%5E2-2ab%2Bb%5E2 pattern. Personally I prefer using the pattern:
+%28x%5E2%29%5E2-2%28x%5E2%29%281%29%2B+1%5E2+=++11x
Simplifying...
+x%5E4-2x%5E2%2B+1+=++11x

3. If there are still square roots...
There are no more square roots so on to step 4.

4. Solve the equation.
The equation is now a 4th degree polynomial equation. To solve this we make one side zero and try to factor the other side. Subtracting 11x from each side:
+x%5E4-2x%5E2-11x%2B+1+=++0
But this equation will not factor. So we will not be able to find solutions with algebra (unless you have learned the very complex formulas for solving 4th degree polynomials). Are you sure you posted the correct equation?

From the graph below, we can see that there are solutions near 0 and 2.5. (They must be irrational solutions since none of the possible rational roots are actually roots.) The best you can do is
  • Plot the graph of y+=+x%5E2-+sqrt%28+11x+%29-+1
  • Use the trace function on a graphing calculator to find the two x coordinates of the points where this graph crosses the x-axis (where the y coordinate is zero).
graph%28800%2C+800%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E4-2x%5E2-11x%2B+1%29