Square Roots: Additional Practice (Set 1)

Find the non-negative number whose square is $225$. Try $15$.

$15^2 = 225 \quad\Rightarrow\quad \sqrt{225} = 15$

The largest perfect square dividing $32$ is $16$, since $32 = 16 \times 2$. Split the radical.

$\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16}\,\sqrt{2} = 4\sqrt{2}$

The $\sqrt{16} = 4$ comes out, and $\sqrt{2}$ stays.

Combine the roots into one with the product rule, then evaluate.

$\sqrt{3} \times \sqrt{12} = \sqrt{3 \times 12} = \sqrt{36} = 6$

The product of the radicands, $36$, is a perfect square, so the answer is the whole number $6$.

The perfect squares around $99$ are $81 = 9^2$ and $100 = 10^2$, so $9 < \sqrt{99} < 10$. Compare distances.

$100 - 99 = 1, \qquad 99 - 81 = 18$

The radicand is far nearer $100$, so $\sqrt{99}$ is closest to $10$.

Compare by squaring. Squaring $5$ gives $25$ and squaring $5.2$ gives $27.04$, while the radicands are $26$ and $24$.

$\sqrt{24} \approx 4.90, \quad 5, \quad \sqrt{26} \approx 5.10, \quad 5.2$

Since $24 < 25$, $\sqrt{24} < 5$, so $\sqrt{24}$ is the smallest.

First simplify $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$, then multiply by the $3$ in front.

$3\sqrt{8} = 3 \times 2\sqrt{2} = 6\sqrt{2}$

The coefficient $3$ multiplies the $2$ to give $6\sqrt{2}$.

A root is rational when the radicand is a perfect square. Here $\sqrt{400} = 20$, $\sqrt{0.09} = 0.3$, and $\sqrt{169} = 13$, but $2$ is not a perfect square.

$1^2 = 1 < 2 < 4 = 2^2$

So $\sqrt{2}$ is irrational; its decimal runs on forever without repeating.

A quotient of square roots splits, so root the numerator and the denominator separately.

$\sqrt{\frac{49}{100}} = \frac{\sqrt{49}}{\sqrt{100}} = \frac{7}{10}$

Check: $\left(\tfrac{7}{10}\right)^2 = \tfrac{49}{100}$.

The side of a square is the square root of its area.

$\text{side} = \sqrt{256} = 16 \text{ cm}$

Since $16^2 = 256$, the side is $16$ centimeters.

Compare by squaring: $\sqrt{8}$ squares to $8$, $3$ squares to $9$, and $\sqrt{10}$ squares to $10$.

$8 < 9 < 10 \quad\Rightarrow\quad \sqrt{8} < 3 < \sqrt{10}$

So the increasing order is $\sqrt{8}, 3, \sqrt{10}$.

Squaring a square root returns the radicand, so $\left(\sqrt{6}\right)^2 = 6$. Then multiply.

$\left(\sqrt{6}\right)^2 \times 3 = 6 \times 3 = 18$

The square undoes the root, leaving $6$ to multiply by $3$.

Combine the roots into one with the quotient rule, then evaluate.

$\sqrt{50} \div \sqrt{2} = \sqrt{50 \div 2} = \sqrt{25} = 5$

The quotient of the radicands, $25$, is a perfect square, so the answer is the whole number $5$.