|
Question 613983: x+√x+5=7
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
The easy way, IMHO, to solve this is to recognize that since the exponent of x, 1, is twice the exponent of  , 1/2, this equation is in what is called "quadratic form". Quadratic form equations can be solved like quadratic equations.
If you are new to equations of quadratic form equations, then it can be very helpful to use a temporary variable:
Let q =
Then 
Substituting these into our equation we get:
This is obviously a quadratic equation. To solve, we first make one side equal to zero. Subtracting 7 from each side we get:
Then we factor:
(q + 2)(q - 1) = 0
One of these factors must be zero:
q + 2 = 0 or q - 1 = 0
Solving these we get:
q = -2 or q = 1
But we are not interested in what values "q" might be. We want to know values for "x". So we substitute back in for the q's:
 or 
Since square roots are never negative there is no solution to the first equation. But we will get solution from the second equation. Squaring both sides we get:
x = 1
Since we squared both sides, we must check our solution. Use the original equation to check:
Checking x = 1:
7 = 7 Check!
After a few of these quadratic form equations, you will no longer need a temporary variable. You will see that
will factor into:
etc.
An alternate way to solve this is as a square root equation:
1. Isolate a the square root
Subtracting x and 5 from each side:
2. Square both sides:
3. Solve the resulting equation.
This is quadratic so we want one side to be zero, Subtracting x from each side:
4. Factor (or use the Quadratic Formula):
(x - 4)(x - 1) = 0
5. One factor must be zero:
x - 4 = 0 or x - 1 = 0
6. Solve
x = 4 or x = 1
7. Check. (Again, we squared both sides earlier so the check is not optional!)
Checking x = 4:
4 + 2 + 5 = 7
11 = 7 Check failed! Reject x = 4
Checking x = 1:
7 = 7 Check!
|
|
|
| |
