Study questions platform-wide or filter by specific tests with correct answers revealed.
Let integer $v > 1$ be a perfect square, so $v = k^2$ for some integer $k \geq 2$, and $\sqrt{v} = k$. Which of the following expressions must also be a perfect square? Select all that apply.
Since $v = k^2$ and $\sqrt{v} = k$:
Option A โ $81v$: $= 81k^2 = (9k)^2$. Always a perfect square. โ
Option B โ $25v + 10\sqrt{v} + 1$: $= 25k^2 + 10k + 1 = (5k+1)^2$. Always a perfect square. โ
Option C โ $4v^2 + 4\sqrt{v} + 1$: $= 4k^4 + 4k + 1$. Test $k = 2$: $4(16) + 4(2) + 1 = 73$. $\sqrt{73}$ is not an integer. Not always a perfect square. โ
Option D โ $v + 2\sqrt{v} + 1$: $= k^2 + 2k + 1 = (k+1)^2$. Always a perfect square. โ
Correct answers: A, B, and D (indices 0, 1, 3).
The concept of 'Rule of Law' was popularized by which British jurist?
Albert Venn Dicey is famous for his exposition of the Rule of Law in his work on the UK Constitution.
Sign in to join the conversation and share your thoughts.
Log In to Comment