How to find the square root - ISEE Middle Level Quantitative Reasoning

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Question

Give the square root of 256.

Answer

$16\cdot16= 256$ - that is, 16 squared is 256, making 16 the square root of 256 by definition.

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Question

Which is the greater quantity?

(a) $6 \sqrt{ 7}$

(b) $7 \sqrt{6}$

Answer

(a) $\left (6 \sqrt{7} \right )^{2} = 6^{2}\left ( \sqrt{7} \right ) ^{2} = 36 \cdot 7 = 252$ , so $6 \sqrt{7} = \sqrt{ 252 }$

(b) $\left (7 \sqrt{6} \right )^{2} = 7^{2}\left ( \sqrt{6} \right ) ^{2} = 49 \cdot 6 = 294$ , so $7\sqrt{6} = \sqrt{ 294 }$

,

so

$\sqrt{294 }> \sqrt{252}$ ,

and

$7\sqrt{6} > 6 \sqrt{7}$

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Question

Which is the greater quantity?

(a) $\sqrt{425 }$

(b)

Answer

$21^{2} = 21 \times 21 = 441$ , so $21 = \sqrt{441}$

, so

$21 = \sqrt{441} > \sqrt{425}$

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Question

Column A Column B

$\sqrt{100}$ $\sqrt{10.0}$

Answer

The square root of 100 is 10, while the square root of 10 is between the square root of 9 and 16, so about 4. Therefore, Column A has to be greater.

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Question

Which is the greater quantity?

(A)

(B) $\sqrt{600}$

Answer

$25 ^{2} = 25 \times 25 = 625$ , so

$25 = \sqrt{625} > \sqrt{600}$

This makes (A) greater.

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Question

Which of the following is equal to $\sqrt{11^{2}+ 6 \cdot 8}$ ?

Answer

$\sqrt{11^{2}+ 6 \cdot 8}$

$= \sqrt{121+ 6 \cdot 8}$

$= \sqrt{121+ 48}$

$= \sqrt{169}$

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Question

$F^{2}= 36$

Which is the greater quantity?

(A)

(B)

Answer

If $F^{2} = 36$ , then one of two things is true - either or .

However, and , so it is impossibe to tell whether (A) or (B) is greater.

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Question

Which of the following is equal to $\sqrt{2^{3}+ 6^{2}+10^{2}}$ ?

Answer

$\sqrt{2^{3}+ 6^{2}+10^{2}}$

$=\sqrt{8+36 +100}$

$=\sqrt{ 144}$

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Question

Which of the following is equal to $\sqrt{5^{2}+7^{2}+7}$ ?

Answer

$\sqrt{5^{2}+7^{2}+7}$

$=\sqrt{25 +49+7}$

$=\sqrt{81}$

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Question

$Q^{2} = 49$

Which is the greater quantity?

(A)

(B)

Answer

If $Q^{2} = 49$ , then one of two things is true - either or . Since and , either way, so (A) is greater.

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Question

$V^{2} = 72$ ; is positive.

Which is the greater quantity?

(a)

(b)

Answer

Since is positive, we can compare to 8 by comparing their squares.

$V^{2} = 72$ and $8^{2} = 8 \times 8 = 64$

$V^{2} > 8^{2}$ , so , making (A) greater.

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Question

Which of the following is equal to $\sqrt{4^{2} \cdot 3^{2}}$ ?

Answer

First, evalutate the terms under the radical:

$\sqrt{4^{2} \cdot 3^{2}} = \sqrt{16 \cdot 9} =\sqrt{144}$

Then, take the square root:

$\sqrt{144}=12$

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Question

Which of the following is equal to $\sqrt{4 \cdot 9 \cdot 16}$ ?

Answer

First, simplify the terms within the square root by multiplying.

$\sqrt{4 \cdot 9 \cdot 16} = \sqrt{36 \cdot 16} = \sqrt{576}$

Then, solve the sqaure root.

$\sqrt{576}=24$

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Question

Which is the greater quantity?

(A) $\sqrt{9} + \sqrt{25}$

(B) $\sqrt{34}$

Answer

$\sqrt{9} + \sqrt{25} = 3 + 5 = 8$

since $8^{2} = 8 \times 8 = 64$ , $8 = \sqrt{64}$ .

Since , $\sqrt{64} > \sqrt{34}$ , and (A) is greater

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Question

Which is the greater quantity?

(A) $\sqrt{4 \cdot 9 \cdot 25}$

(B) $2 \cdot 3 \cdot 5$

Answer

$\sqrt{4 \cdot 9 \cdot 25} = \sqrt{900} = 30$

$2 \cdot 3 \cdot 5 = 30$

The quantities are equal.

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Question

Which is the greater quantity?

(A) $\sqrt{1+ 4+ 9 + 16 + 25 + 36}$

(B)

Answer

Therefore, $\sqrt{1+ 4+ 9 + 16 + 25 + 36} = \sqrt{91}$ .

Since $9^{2} = 9 \times 9 = 81$ , $9 = \sqrt{81}$ .

, so $\sqrt{91 }>\sqrt{ 81}$ , and (A) is greater.

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Question

Which is the greater quantity?

(A) $\sqrt{100^{4}}$

(B) $100^{2}$

Answer

$\sqrt{100^{4}} = \sqrt{100 \times 100 \times 100 \times 100} = \sqrt{100,000,000} = 10,000$

$100^{2} = 100 \times 100 = 10,000$

The quantities are equal.

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Question

Which of the following is equal to 39?

Answer

$12\cdot3+3$ is equal to:

Therefore, $12\cdot3+3$ is the correct answer.

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Question

Simplify the below:

$\sqrt{24}$

Answer

When breaking down a radical, we first want to find the largest perfect square that might be a factor for the number under the radical.

We start with 4, 9, 16, 25 etc. until we find which one is a factor.

In this case, 4 is a factor of 24.

We can now break down the radical to become:

$\sqrt{24}=\sqrt4\sqrt6$

The square root of 4 becomes 2 and the square root of 6 will not break down any further, this leads us to the answer below:

$\sqrt{24}=\sqrt4\sqrt6=2\sqrt6$

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Question

Simplify:

$2\sqrt{72}$

Answer

When breaking down a square root, we must first find the largest perfect square factor that goes into the number under the radical; starting with 4, 9, 16, 25, 36 etc.

In this case, 36 will go into 72, 2 times.

Which reduces the radical to the below:

$2\sqrt{72}=2(\sqrt{36}\sqrt2)$

We can then simplify square root 36 to become 6 and we get:

$2\sqrt{72}=2(\sqrt{36}\sqrt2)=2(6\sqrt2)$

When we multiply with a radical, only the numbers outside the radical are multiplied.

$2(6\sqrt2)=12\sqrt2$

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