Square Roots: Practice

Evaluate each root separately, then add. A root does not combine across addition, so each is found on its own first.

$\sqrt{36} + \sqrt{64} = 6 + 8 = 14$

Note this is not $\sqrt{36 + 64} = \sqrt{100} = 10$; the sum of the roots differs from the root of the sum.

The largest perfect square dividing $200$ is $100$, since $200 = 100 \times 2$. Split the radical.

$\sqrt{200} = \sqrt{100 \times 2} = \sqrt{100}\,\sqrt{2} = 10\sqrt{2}$

The $\sqrt{100} = 10$ comes out, and $\sqrt{2}$ stays.

Compare by squaring each. Squaring $7$ gives $49$ and squaring $6.5$ gives $42.25$, while the radicands are $50$ and $40$.

$\sqrt{50} \approx 7.07, \quad 7, \quad \sqrt{40} \approx 6.32, \quad 6.5$

Since $50 > 49$, $\sqrt{50} > 7$, so $\sqrt{50}$ is the largest.

Squaring a square root returns the radicand, so $\left(\sqrt{11}\right)^2 = 11$. Then add.

$\left(\sqrt{11}\right)^2 + 4 = 11 + 4 = 15$

The square undoes the root even when the radicand is not a perfect square.

Evaluate each root, then multiply (or use $\sqrt{a}\,\sqrt{b} = \sqrt{ab}$).

$\sqrt{16} \times \sqrt{9} = 4 \times 3 = 12$

Equivalently $\sqrt{16 \times 9} = \sqrt{144} = 12$.

Add inside the radical first, then take the root of the single number.

$\sqrt{49 + 576} = \sqrt{625} = 25$

Since $25^2 = 625$, the root is exactly $25$. It is not $\sqrt{49} + \sqrt{576} = 7 + 24 = 31$.

The largest perfect square dividing $98$ is $49$, since $98 = 49 \times 2$. Split the radical.

$\sqrt{98} = \sqrt{49 \times 2} = \sqrt{49}\,\sqrt{2} = 7\sqrt{2}$

The $\sqrt{49} = 7$ comes out, and $\sqrt{2}$ stays.

The perfect squares around $7$ are $4 = 2^2$ and $9 = 3^2$.

$4 < 7 < 9 \quad\Rightarrow\quad 2 < \sqrt{7} < 3$

So $\sqrt{7}$ lies between $2$ and $3$. It is irrational, so it is not $3.5$ and not a whole number, and it is less than $3$.

Both terms share the same radical part $\sqrt{2}$, so they add like $5$ of a thing plus $3$ of the same thing.

$5\sqrt{2} + 3\sqrt{2} = (5 + 3)\sqrt{2} = 8\sqrt{2}$

The $\sqrt{2}$ stays as the common factor; only the coefficients add.

If $\sqrt{n} = 14$, then $n$ is what you get by squaring $14$, because squaring undoes the root.

$n = 14^2 = 196$

Check: $\sqrt{196} = 14$.

A root is rational when the radicand is a perfect square (of a whole number or a tidy decimal). Here $\sqrt{121} = 11$, $\sqrt{0.04} = 0.2$, and $\sqrt{225} = 15$, but $12$ is not a perfect square.

$3^2 = 9 < 12 < 16 = 4^2$

So $\sqrt{12}$ is irrational; its decimal never ends or repeats.

The side of a square is the square root of its area, so root the numerator and the denominator.

$\text{side} = \sqrt{\tfrac{1}{4}} = \frac{\sqrt{1}}{\sqrt{4}} = \frac{1}{2} \text{ ft}$

Check: $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$, the area.