Forward Grade 8 The Number System. Practice it free.

$5^2 = 25 < 27 < 36 = 6^2$, so $\sqrt{27}$ is between $5$ and $6$; the larger is $6$.

Multiply coefficients ($2 \cdot 3 = 6$) and add exponents: $3 + 4 = 7\textsf{.}$ So $(2 \times 10^{3})(3 \times 10^{4}) = 6 \times 10^{7}\textsf{.}$

$9 \times 9 = 81$ and $(-9) \times (-9) = 81\textsf{,}$ so $x = \pm 9\textsf{.}$

Take the positive square root: $x = \sqrt{169} = 13$ because $13 \times 13 = 169$.

$9^2 = 81$ is very close to $83$ and $\sqrt{83} \approx 9.1$, so the closest integer is $9$.

Divide coefficients ($9 \div 3 = 3$) and subtract exponents: $8 - 2 = 6\textsf{.}$ So $\dfrac{9 \times 10^{8}}{3 \times 10^{2}} = 3 \times 10^{6}\textsf{.}$

$8 \times 8 = 64$ and $(-8) \times (-8) = 64\textsf{,}$ so $x = \pm 8\textsf{.}$

$12 \times 12 = 144$, so $\sqrt{144} = 12$.