SOLUTION: Find all real solutions of the equation x-7sqrt(x) +10=0 10 is not in the sqrt what steps must I first take to get rid of the square root

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Question 779265: Find all real solutions of the equation
x-7sqrt(x) +10=0
10 is not in the sqrt
what steps must I first take to get rid of the square root
Found 3 solutions by Edwin McCravy, AnlytcPhil, jsmallt9:
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
There are two ways to do this problem.  The way you're probably
studying now is u-substitution:

       x-7√x+10 = 0

Let u = √x, then u² = x

       u²-7u+10 = 0

     (u-2)(u-5) = 0

  u-2 = 0;  u-5 = 0
    u = 2     u = 5

   √x = 2;     √x = 5 
 
(√x)² = 2²; (√x)² = 5²
    x = 4;      x = 25

We must always check radical
equations to find out if either
or both of those answers are 
legitimate solutions or are
only extraneous.

Checking x = 4:

       x-7√x+10 = 0
       4-7√4+10 = 0
       4-7·2+10 = 0
        4-14+10 = 0
              0 = 0

So 4 is a legitimate solution.

Checking x = 25:

       x-7√x+10 = 0
      25-7√25+10 = 0
      25-7·5+10 = 0
       25-35+10 = 0
              0 = 0 

So 25 is also a legitimate solution.

Edwin

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Here's the other way to solve that problem:

     x-√x+10 = 0

        x+10 = 7√x

     (x+10)² = (7√x)²

(x+10)(x+10) = (7√x)²

  x²+20x+100 = 49x
  x²-29x+100 = 0
 (x-25)(x-4) = 0

  x-25 = 0;  x-4 = 0
     x = 25;   x = 4

Checking is the the same as in the other way to solve it.

Edwin

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The solution provided by another tutor is perfectly correct. But the "what steps ... get rid of the square root" in your post suggests that this other solution does not use the method you may be learning in school.

To eliminate a square root:
  1. Isolate that square root term on one side of the equation.
  2. Square both sides

x-7sqrt%28x%29%2B10=0
Adding the square root term we get:
x%2B10=7sqrt%28x%29
(The square root term is sufficiently isolated. If the 7 in front bothers you, then you can divide both sides by 7.) Squaring both sides:
%28x%2B10%29%5E2=%287sqrt%28x%29%29%5E2
x%5E2%2B20x%2B100=49x

With the square root eliminated we can now solve the resulting equation. This is quadratic so we want a zero on one side. Subtracting 49x from each side we get:
x%5E2-29x%2B100=0
Factoring:
%28x-25%29%28x-4%29=0
From the Zero Product Property:
x-25 = 0 or x-4 = 0
which give us:
x = 25 or x = 4

And just as the other tutor said, we must check our solutions. Both of them work so they are both solutions to the original equation.