SOLUTION: Find all the real solutions of the equation by first rewriting the equation as a quadratic equation. (Enter your answers as a comma-separated list.) 7x − 37Squareroot of

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Question 623062: Find all the real solutions of the equation by first rewriting the equation as a quadratic equation. (Enter your answers as a comma-separated list.)

7x − 37Squareroot of x + 48 = 0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Please put radicands in parentheses. ("Radicand" is the name for the expression inside a radical.) The way you posted the problem it is not possible to know if it is:
7x-37sqrt%28x%2B48%29+=+0
or
7x-37sqrt%28x%29%2B48+=+0

Since I don't know which one is real, I'm just going to get you started on both and tell you how to finish them.

7x-37sqrt%28x%2B48%29+=+0
Isolate the square root by adding it to both sides:
7x+=+37sqrt%28x%2B48%29
Then
  1. Square both sides. This will eliminate the square root.
  2. Solve the equation. (It will be a quadratic equation.)
  3. Check each solution. This is not optional since you've squared both sides of the equation. Use the original equation to check and reject any "solutions" that do not check out.

7x-37sqrt%28x%29%2B48+=+0
This one can be solved in a similar way to the one above. But there is a faster way. Since the exponent on x in 7x is 1 and the exponent on x in 37sqrt%28x%29 is 1/2 (Remember that square roots can be expressed as exponents of 1/2?) Since the exponent on x is twice as large as the exponent on sqrt%28x%29, this equation is in what is called quadratic form. It is not literally a quadratic equation but it has the same pattern of exponents. And we can solve this in a similar way.

Exactly like
7q%5E2-37q%2B48=0
can be factored into
%287q-16%29%28q-3%29+=+0
7x-37sqrt%28x%29%2B48+=+0
can be factored into
%287sqrt%28x%29-16%29%28sqrt%28x%29-3%29+=+0
We can now use the Zero Product Property which tells us that one of these factors must be zero:
7sqrt%28x%29-16+=+0 or sqrt%28x%29-3+=+0
Solving these...
7sqrt%28x%29+=+16
sqrt%28x%29+=+16%2F7
%28sqrt%28x%29%29%5E2+=+%2816%2F7%29%5E2
x+=+256%2F49

sqrt%28x%29-3+=+0
sqrt%28x%29=+3
%28sqrt%28x%29%29%5E2=+%283%29%5E2
x+=+9

Just like the other problem, we must check our answers because we squared both sides of the equation. I'll leave that up to you.