The Pyth. Thm. applies here:
(√x + 1)^2 + (2√x)^2 = (2√x + 1 )^2
Expanding the squares:
x + 2sqrt(x) + 1 + 4x = 4x + 4sqrt(x) + 1
Let's subtract x + 2sqrt(x) + 1 + 4x from both sides:
4x + 4sqrt(x) + 1
-(x + 2sqrt(x) + 1 + 4x)
-------------------------------
3x + 2sqrt(x) - 4x = 0
Then 2sqrt(x) = x
Squaring both sides, 4x = x^2, or x^2 - 4x = 0. Then (x-4)x = 0, and the two possible solutions are 0 and 4.
Check these results by substitution. Does the Pyth. Thm. hold true for x=4?
Answer: -8 and -4
This is something you do through trial and error. Making a list or a table like shown below might help.
12 years old, cause when u multiply 12 times 4 u get 48 and 48 + 12= 57
Answer:
4
Step-by-step explanation:
4
