Square Roots and Radicals - GED Math

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Question

Which of the following is true of $\sqrt{792}$ ?

Answer

To determine which two consecutive integers flank $\sqrt{792}$ out of the set $\left \{ 27, 28, 29, 30, 31\right \}$ , square each number. The squares of each number in the set are:

$27^{2} = 27 \times 27 = 729$

$28^{2}= 28 \times 28 = \textbf{784}$

$29^{2} = 29 \times 29 = \textbf{841}$

$30^{2} = 30 \times 30 = 900$

$31^{2} = 31 \times 31 = 961$

;

it follows that

$\sqrt{784 }< \sqrt{729 }< \sqrt{841}$

or

$28 < \sqrt{729 }<29$ .

This is the correct choice.

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Question

Which of the following is true of $\sqrt{792}$ ?

Answer

To determine which two consecutive integers flank $\sqrt{792}$ out of the set $\left \{ 27, 28, 29, 30, 31\right \}$ , square each number. The squares of each number in the set are:

$27^{2} = 27 \times 27 = 729$

$28^{2}= 28 \times 28 = \textbf{784}$

$29^{2} = 29 \times 29 = \textbf{841}$

$30^{2} = 30 \times 30 = 900$

$31^{2} = 31 \times 31 = 961$

;

it follows that

$\sqrt{784 }< \sqrt{729 }< \sqrt{841}$

or

$28 < \sqrt{729 }<29$ .

This is the correct choice.

Compare your answer with the correct one above

Question

Which of the following is true of $\sqrt{792}$ ?

Answer

To determine which two consecutive integers flank $\sqrt{792}$ out of the set $\left \{ 27, 28, 29, 30, 31\right \}$ , square each number. The squares of each number in the set are:

$27^{2} = 27 \times 27 = 729$

$28^{2}= 28 \times 28 = \textbf{784}$

$29^{2} = 29 \times 29 = \textbf{841}$

$30^{2} = 30 \times 30 = 900$

$31^{2} = 31 \times 31 = 961$

;

it follows that

$\sqrt{784 }< \sqrt{729 }< \sqrt{841}$

or

$28 < \sqrt{729 }<29$ .

This is the correct choice.

Compare your answer with the correct one above

Question

Which of the following is true of $\sqrt{792}$ ?

Answer

To determine which two consecutive integers flank $\sqrt{792}$ out of the set $\left \{ 27, 28, 29, 30, 31\right \}$ , square each number. The squares of each number in the set are:

$27^{2} = 27 \times 27 = 729$

$28^{2}= 28 \times 28 = \textbf{784}$

$29^{2} = 29 \times 29 = \textbf{841}$

$30^{2} = 30 \times 30 = 900$

$31^{2} = 31 \times 31 = 961$

;

it follows that

$\sqrt{784 }< \sqrt{729 }< \sqrt{841}$

or

$28 < \sqrt{729 }<29$ .

This is the correct choice.

Compare your answer with the correct one above

Question

Which of the following is true of $\sqrt{792}$ ?

Answer

To determine which two consecutive integers flank $\sqrt{792}$ out of the set $\left \{ 27, 28, 29, 30, 31\right \}$ , square each number. The squares of each number in the set are:

$27^{2} = 27 \times 27 = 729$

$28^{2}= 28 \times 28 = \textbf{784}$

$29^{2} = 29 \times 29 = \textbf{841}$

$30^{2} = 30 \times 30 = 900$

$31^{2} = 31 \times 31 = 961$

;

it follows that

$\sqrt{784 }< \sqrt{729 }< \sqrt{841}$

or

$28 < \sqrt{729 }<29$ .

This is the correct choice.

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Question

Simplify:
$\small \sqrt{192}$

Answer

$\small \sqrt{192}=\sqrt{64}\times \sqrt3=8\times \sqrt3=8\sqrt3$

An alternate solution is:

$\small \sqrt{192}=\sqrt{16}\times \sqrt{12}=4\times \sqrt{4} \times \sqrt3=4\times2\times\sqrt3=8\sqrt3$

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Question

Simplify:

$\small \sqrt{x^4}$

Answer

$\small \sqrt{x^4}=\sqrt{x^2}\times \sqrt{x^2}=x\times x=x^2$

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Question

Simplify:

$\small \sqrt{98}$

Answer

$\small \sqrt{98}=\sqrt{49}\times \sqrt{2}=7\times \sqrt2=7\sqrt2$

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Question

Simplify:

$\small \sqrt{72}=$

Answer

$\small \sqrt{72}=\sqrt{36}\times \sqrt2=6\times \sqrt2=6\sqrt2$

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Question

Simplify:

$\sqrt{147}$

Answer

$\sqrt{147}=\sqrt{49}\times \sqrt3=7\times \sqrt3=7\sqrt3$

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Question

Refer to the above number line. What point most likely represents the square root of 280?

Do not use a calculator.

Answer

$16^{2} = 256 < 280 < 289 < 17^{2}$

Therefore, $16 < \sqrt{280} < 17$ ,

making the correct choice .

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Question

Simplify:

$\sqrt{42}$

Do not use a calculator.

Answer

The prime factorization of 42 is

$42 = 2 \times 3 \times 7$ .

Since 42 is the product of distinct primes, it has no perfect square factors, and, therefore, its square root cannot be simplified further. It is already in simplifed form.

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Question

Simplify:

$\sqrt{48}$

Do not use a calculator.

Answer

The prime factorization of 48 is

$48 = 2 \times 2 \times 2 \times 2 \times 3$ .

Rewrite, and use the product of radicals property to simplify:

$\sqrt{48}$

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

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

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

$=4 \sqrt{ 3}$

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Question

Factor: $2\sqrt{120}$

Answer

In order to factor the radical, we will need to rewrite 120 as multiples of perfect squares.

$2\sqrt{120} = 2\sqrt{30} \cdot \sqrt{4}$

Reduce the known term

$2\sqrt{30} \cdot \sqrt{4} =2\sqrt{30} \cdot 2 = 4\sqrt{30}$

The value of $\sqrt{30}$ cannot be factored any further.

The answer is: $4\sqrt{30}$

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Question

Factor: $6\sqrt{40}$

Answer

Rewrite the root 40 in factors of perfect squares.

$6\sqrt{40} = 6 \cdot \sqrt{4}\cdot \sqrt{10} = 6 \cdot 2\cdot \sqrt{10}$

The answer is: $12\sqrt{10}$

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Question

Factor: $3\sqrt{88}$

Answer

In order to factor this, we will need to rewrite root 88 by factoring using values of perfect squares.

$3\sqrt{88} = 3\times\sqrt{4}\times \sqrt{22} = 3\times 2 \times \sqrt{22}$

The answer is: $6\sqrt{22}$

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Question

Rationalize: $\frac{3}{2\sqrt2}$

Answer

In order to rationalize the radical, we will need to multiply both the top and bottom by square root two.

$\frac{3}{2\sqrt2}\cdot\frac{\sqrt2}{\sqrt2} = \frac{3\sqrt2}{2\cdot 2}$

The answer is: $\frac{3\sqrt{2}}{4}$

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Question

Rationalize: $\frac{2}{\sqrt5}$

Answer

In order to eliminate the radical on the denominator, we will need to multiply root five on the top and bottom of the fraction.

$\frac{2}{\sqrt5}\cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

The answer is: $\frac{2\sqrt{5}}{5}$

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Question

Rationalize: $\frac{\sqrt3}{\sqrt{20}}$

Answer

Simplify the denominator by factoring using perfect squares.

$\sqrt{20} = \sqrt4 \cdot \sqrt5 = 2\sqrt{5}$

Rewrite the fraction.

$\frac{\sqrt3}{\sqrt{20}}=\frac{\sqrt3}{2\sqrt{5}}$

Multiply the top and bottom by root 5.

$\frac{\sqrt3}{2\sqrt{5}} \cdot \frac{\sqrt5}{\sqrt5} = \frac{\sqrt{15}}{10}$

The answer is: $\frac{\sqrt{15}}{10}$

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Question

Rationalize: $\frac{2}{\sqrt{12}}$

Answer

Factor the denominator by factors of perfect squares.

$\sqrt{12} = \sqrt4 \cdot \sqrt3 = 2\sqrt{3}$

Replace the term.

$\frac{2}{\sqrt{12}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt3}$

Multiply by root three on the top and bottom.

$\frac{1}{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} = \frac{\sqrt3}{3}$

The answer is: $\frac{\sqrt3}{3}$

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