Find the only value of $x$ that satisfies: $$\sqrt{7+\sqrt{4-\sqrt{3+x}}}=3$$
Solve for x:
sqrt(sqrt(4 - sqrt(x + 3)) + 7) = 3
Raise both sides to the power of two:
sqrt(4 - sqrt(x + 3)) + 7 = 9
Subtract 7 from both sides:
sqrt(4 - sqrt(x + 3)) = 2
Raise both sides to the power of two:
4 - sqrt(x + 3) = 4
Subtract 4 from both sides:
-sqrt(x + 3) = 0
Multiply both sides by -1:
sqrt(x + 3) = 0
Square both sides:
x + 3 = 0
Subtract 3 from both sides:
x = -3
+267 If I understood correctly (the $ signs are cluttering up the syntax somewhat) you want to solve the following:
\(\sqrt{7+\sqrt{4-\sqrt{3+x}}}=3\)
We also know that there is only one solution, since you said "only value of x".
One way to solve this is to look at the most outer square root with the 7. We know that the square root of 9 is 3. The only possible way this can hold true is if 7 + the other value is equal to 2. Therefore this has to hold true:
\(\sqrt{4-\sqrt{3+x}}=2\)
By using the same logic, 4 + something has to be equal to 4 in order for its root to be equal to 2. Therefore we can see this has to be true:
\(\sqrt{3+x}=0\)
This is clearly true only if x = -3
You can verify this by putting that value back into the original question. The other more general way is to just square both sides and move the term from the left to the right and keep squareing til you get x alone on the left and a number on the right. This will give you the same result but I thought that the explanation above would give you a better intuition(I hope)
