SOLUTION: (7+(48^1/2))^1/2 + (7-(48^1/2))^1/2

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Question 35597: (7+(48^1/2))^1/2 + (7-(48^1/2))^1/2
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
First, let me make sure I have the problem written correctly in an algebra.com equation:


Also, let me congratulate you on asking one of the hairiest questions I've seen lately. It will be interesting to see if anyone else has an easier solution to this!!!

Let's begin by changing this to radical form, which should be a little easier to read: sqrt%287%2B+sqrt+%2848%29%29%2Bsqrt%287-sqrt%2848%29%29+

Let +x+=sqrt%287%2B+sqrt+%2848%29%29%2Bsqrt%287-sqrt%2848%29%29+ , and square both sides of the equation.


Now, FOIL out this product of two binomials:


Notice that the two sqrt%2848%29 terms subtract out, the 7+ 7 combines to give you 14, and the two sqrt%287-sqrt%2848%29%29%2A+sqrt%287%2Bsqrt%2848%29%29+ terms combine to give you +2sqrt%287-sqrt%2848%29%29%2A+sqrt%287%2Bsqrt%2848%29%29+ .
+x%5E2+=+14%2B+2%2Asqrt%287-sqrt%2848%29%29%2A+sqrt%287%2Bsqrt%2848%29%29+

Also, notice that the two square roots can be multiplied together and placed inside one square root:
+x%5E2+=+14%2B++2%2Asqrt%28%287-sqrt%2848%29%29%2A+%287%2Bsqrt%2848%29%29%29+

Next, FOIL inside the square root:
+x%5E2+=+14%2B+2%2A+sqrt%2849-48%29+

x%5E2+=+14+%2B+2%2A+sqrt%281%29+
x%5E2+=+14+%2B+2
x%5E2+=+16
x=4 or x=+-4

Since the original problem was the sum of two radical expressions, you must reject the negative value. The final answer is x=4, which is a rational number.

R^2 at SCC