Square Roots: Practice

Find the perfect squares on either side of $20$. The square $16 = 4^2$ is just below, and $25 = 5^2$ is just above.

$16 < 20 < 25 \quad\Rightarrow\quad 4 < \sqrt{20} < 5$

So $\sqrt{20}$ lies between $4$ and $5$.

The perfect squares around $37$ are $36 = 6^2$ and $49 = 7^2$, so $6 < \sqrt{37} < 7$. Since $37$ is only just above $36$, the root is very close to $6$.

$37 - 36 = 1, \qquad 49 - 37 = 12$

The radicand is far nearer $36$, so $\sqrt{37}$ is closest to $6$.

The largest perfect square dividing $8$ is $4$, since $8 = 4 \times 2$. Split the radical with the product rule.

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4}\,\sqrt{2} = 2\sqrt{2}$

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

Work out what is inside the radical first, then take the root of the single number.

$\sqrt{9 + 16} = \sqrt{25} = 5$

This is not $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$: a root does not split across a sum.

Evaluate each root, then multiply, or combine first with $\sqrt{a}\,\sqrt{b} = \sqrt{ab}$.

$\sqrt{4}\times\sqrt{25} = 2 \times 5 = 10$

Equivalently $\sqrt{4 \times 25} = \sqrt{100} = 10$.

The largest perfect square dividing $45$ is $9$, since $45 = 9 \times 5$. Split the radical.

$\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9}\,\sqrt{5} = 3\sqrt{5}$

The $\sqrt{9} = 3$ comes out, and $\sqrt{5}$ stays.

Find the non-negative number whose square is $0.25$. Try $0.5$.

$0.5^2 = 0.5 \times 0.5 = 0.25 \quad\Rightarrow\quad \sqrt{0.25} = 0.5$

A square root can apply to a decimal too; it is still the number that squares to the radicand.

First take the principal root, then apply the minus sign that sits in front of it.

$\sqrt{49} = 7, \qquad -\sqrt{49} = -7$

The radical itself is the non-negative $7$; the leading minus makes the whole expression $-7$.

Since $2$ is not a perfect square, $\sqrt{2}$ is irrational, so its decimal never ends and never repeats.

$\sqrt{2} = 1.41421356\ldots$

Values like $1.41$ or $\tfrac{7}{5} = 1.4$ are only approximations, not the exact value.

Find the non-negative number whose square is $169$. Try $13$.

$13^2 = 169 \quad\Rightarrow\quad \sqrt{169} = 13$

The perfect squares around $102$ are $100 = 10^2$ and $121 = 11^2$, so $10 < \sqrt{102} < 11$. Since $102$ is only just above $100$, the root is close to $10$.

$102 - 100 = 2, \qquad 121 - 102 = 19$

The radicand is far nearer $100$, so $\sqrt{102}$ is closest to $10$.

Like a product, a quotient of square roots splits, so root the numerator and the denominator separately.

$\sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}$

Check by squaring: $\left(\tfrac{3}{4}\right)^2 = \tfrac{9}{16}$.