Question

Blank radical 12 + 2 radical 27 = 14 radical blanks

Answer 1

Let's represent the problem as an equation to solve for the blank radical.

First, we need to simplify the terms:

1. Simplify \(\sqrt{12}\):
\[
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}
\]

2. Simplify \(\sqrt{27}\):
\[
\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}
\]

Now, substituting these simplifications back into the equation gives us:
\[
x\sqrt{12} + 2\sqrt{27} = 14\sqrt{y}
\]
\[
x(2\sqrt{3}) + 2(3\sqrt{3}) = 14\sqrt{y}
\]
\[
2x\sqrt{3} + 6\sqrt{3} = 14\sqrt{y}
\]

Now, factor out \(\sqrt{3}\) on the left side:
\[
\sqrt{3}(2x + 6) = 14\sqrt{y}
\]

Now, we can equate the coefficients of \(\sqrt{3}\) from both sides. This leads us to:
\[
2x + 6 = 14\sqrt{y}/\sqrt{3}
\]

To continue, we need \(y\) to resolve this comparison. Assuming \(y = 3\) is a potential blank that could lead to integer simplification:
\[
14\sqrt{3}/\sqrt{3} = 14
\]
Thus:
\[
2x + 6 = 14
\]
Solving for \(x\):
\[
2x = 14 - 6
\]
\[
2x = 8 \implies x = 4
\]

Thus, we find that filling in the blanks gives us:
\[
4\sqrt{12} + 2\sqrt{27} = 14\sqrt{3}
\]

Finally, the values for the blanks are \(4\) and \(3\):
\[
4 \quad \text{and} \quad 3
\]