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(\sqrt(3)+\sqrt(7))^(2)
![Snowy days ❄⛄](/images/users/0/1/128x128.jpeg)
2 years ago
![Snowy days ❄⛄](/images/users/0/1/128x128.jpeg)
2 years ago
Btw im a girl :) just to clarify
![oobleck](/images/users/0/1/128x128.jpeg)
2 years ago
(√3 + √7)^2 = (√3 + √7)(√3 + √7) = 3 + 2√21 + 7
![Snowy days ❄⛄](/images/users/0/1/128x128.jpeg)
2 years ago
You cared enough to respond, so you obviously care.
![:)](/images/users/0/1/128x128.jpeg)
2 years ago
The post above Snowy days was: No body cares snowy.
Then she replied: You cared enough to reply, so you obviously care.
![:)](/images/users/0/1/128x128.jpeg)
2 years ago
Since it was deleted.
![Explain Bot](/images/users/4931/128x128.jpeg)
7 months ago
To simplify the expression (\sqrt(3)+\sqrt(7))^(2), we can use the formula for expanding the square of a binomial:
(a + b)^2 = a^2 + 2ab + b^2.
In this case, a = \sqrt(3) and b = \sqrt(7). Substituting these values into the formula, we get:
(\sqrt(3)+\sqrt(7))^(2) = (\sqrt(3))^2 + 2(\sqrt(3))(\sqrt(7)) + (\sqrt(7))^2.
Simplifying further:
(\sqrt(3))^2 = 3,
(\sqrt(7))^2 = 7,
2(\sqrt(3))(\sqrt(7)) = 2√(3)√(7).
Since the square of a square root (√x)^2 is simply x, we get:
3 + 2√(3)√(7) + 7.
Combining like terms, we have:
10 + 2√(21).
Therefore, (\sqrt(3)+\sqrt(7))^(2) simplifies to 10 + 2√(21).