Square Roots and Operations

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PSAT Math › Square Roots and Operations

Questions 1 - 10

Simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$-\sqrt{3}+9$

$\sqrt{63}$

$2\sqrt{3}$

$7\sqrt{3}+9$

Explanation

To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow \sqrt{9*3}\rightarrow \sqrt{3^2*3}\rightarrow 3\sqrt{3}$

$\sqrt{48}\rightarrow \sqrt{16*3}\rightarrow \sqrt{4^2*3}\rightarrow 4\sqrt{3}$

$\sqrt{81}\rightarrow\sqrt{9*9}\rightarrow\sqrt{9^2}\rightarrow9$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

Simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$-\sqrt{3}+9$

$\sqrt{63}$

$2\sqrt{3}$

$7\sqrt{3}+9$

Explanation

To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow \sqrt{9*3}\rightarrow \sqrt{3^2*3}\rightarrow 3\sqrt{3}$

$\sqrt{48}\rightarrow \sqrt{16*3}\rightarrow \sqrt{4^2*3}\rightarrow 4\sqrt{3}$

$\sqrt{81}\rightarrow\sqrt{9*9}\rightarrow\sqrt{9^2}\rightarrow9$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

If $\sqrt{x}=3^2$ what is $x$ ?

$81$

$9$

$3$

$729$

$27$

Explanation

Square both sides:

x = (32)2 = 92 = 81

If $\sqrt{x}=3^2$ what is $x$ ?

$81$

$9$

$3$

$729$

$27$

Explanation

Square both sides:

x = (32)2 = 92 = 81

The square root(s) of 36 is/are ________.

6 and -6

-6

6, -6, and 0

None of these answers are correct.

Explanation

To square a number is to multiply that number by itself. Because 6 x 6 = 36 AND -6 x -6 = 36, both 6 and -6 are square roots of 36.

The square root(s) of 36 is/are ________.

6 and -6

-6

6, -6, and 0

None of these answers are correct.

Explanation

To square a number is to multiply that number by itself. Because 6 x 6 = 36 AND -6 x -6 = 36, both 6 and -6 are square roots of 36.

Evaluate: $\sqrt{108}-\sqrt{81}$

$6\sqrt{3}-9$

$24\sqrt{3}-81$

$24\sqrt{3}-9$

$4\sqrt{27}-\sqrt{9}$

None of the available answers

Explanation

Let us factor 108 and 81

$108=2\times 54= 2\times 2\times 27= 2^2\times 3\times 9= 2^2\times 3^2\times 3$

$81=9\times 9=9^2$

$\sqrt{108}-\sqrt{81}=6\sqrt{3}-9$

Evaluate: $\sqrt{108}-\sqrt{81}$

$6\sqrt{3}-9$

$24\sqrt{3}-81$

$24\sqrt{3}-9$

$4\sqrt{27}-\sqrt{9}$

None of the available answers

Explanation

Let us factor 108 and 81

$108=2\times 54= 2\times 2\times 27= 2^2\times 3\times 9= 2^2\times 3^2\times 3$

$81=9\times 9=9^2$

$\sqrt{108}-\sqrt{81}=6\sqrt{3}-9$

Simplify:

$\sqrt{\frac{64}{169}}$

$\frac{8}{13}$

$\frac{64}{169}$

$\frac{8}{15}$

$\frac{7}{13}$

Explanation

To simplfy, we must first distribute the square root.

$\frac{\sqrt{64}}{\sqrt{169}}$

Next, we can simplify each of the square roots.

$\frac{8}{13}$

Simplify:

$\sqrt{\frac{64}{169}}$

$\frac{8}{13}$

$\frac{64}{169}$

$\frac{8}{15}$

$\frac{7}{13}$

Explanation

To simplfy, we must first distribute the square root.

$\frac{\sqrt{64}}{\sqrt{169}}$

Next, we can simplify each of the square roots.

$\frac{8}{13}$

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