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

Card 0 of 1665

Question

$72- \sqrt{64}=$

Answer

First, find the square root:

$\sqrt{64}=8$

Then, solve:

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Question

$72* \sqrt{64}=$

Answer

First, find the square root:

$\sqrt{64}=8$

Then, solve:

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Question

Solve the equation:

$b^{2} =\frac{9}{441}$

Answer

$b^{2} =\frac{9}{441}$

Take the square root of both sides of the equation:

$\sqrt{b^{2}} =\sqrt{\frac{9}{441}}$

For the equation $\sqrt{b^{2}} =\sqrt{\frac{9}{441}}$ there are 2 solutions, because it is positive: $y=\sqrt{a}$ , which is positive, and $y = -\sqrt{a}$ , which is negative. Together, these two roots are denoted $\pm\sqrt{a}$ .

$3\times3 = 9$

$-3 \times -3 = 9$

$-21\times-21 =441$

$21\times21 = 441$

$b = \frac{3}{21} or -\frac{3}{21}$

$b=\pm\frac{1}{7}$

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Question

$\sqrt{169}=$

Answer

Find the square root:

$\sqrt{169}=13$

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Question

$\sqrt{196}=$

Answer

Find the square root:

$\sqrt{196}=14$

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Question

$\sqrt{225}=$

Answer

Find the square root:

$\sqrt{225}=15$

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Question

$\sqrt{64}=$

Answer

Find the square root of 64:

$\sqrt{64}=8$

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Question

$\sqrt{400}=$

Answer

Find the square root:

$\sqrt{400}=20$

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Question

$\sqrt{100}=$

Answer

Find the square root: $\sqrt{100}=10$

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Question

Estimate $\sqrt{82}$ to the nearest whole number.

Answer

The first perfect square less than 82 is 81.

The first perfect square greater than 82 is 100.

$\sqrt{81} < \sqrt{82} < \sqrt{100}$

$9< \sqrt{82} < 10$

$\sqrt{82}$ is between and

Since is closer to than to the best whole number estimate for

$\sqrt{82}$ is

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Question

Estimate to the nearest whole number.

$\sqrt{14.3}$

Answer

$\sqrt{14.3}$

The first perfect square less than is .

The first perfect square greater than is

$\sqrt{9}< \sqrt{14.3} < \sqrt{16}$

$3< \sqrt{14.3} < 4$

Therefore the $\sqrt{14.3}$ lies between and

Because 14.3 is closer to 16, the best whole number estimate for $\sqrt{14.3}$ is

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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

Estimate $y^{2} =40$ to the nearest whole integer.

Answer

$y^{2} =40$

Take the square root of both sides of the equation.

$\sqrt{y^{2}} = \sqrt{40}$

The first perfect square that is less than is

The first perfect square that is greater than is

$\sqrt{36} < \sqrt{40} < \sqrt{49}$

$6< \sqrt{40} < 7$

The $\sqrt{40}$ ls between and

is closer to than .

For the equation there are 2 solutions, because is positive: $y=\sqrt{a}$ , which is positive, and $y = -\sqrt{a}$ , which is negative. Together, these two roots are denoted $\pm\sqrt{a}$ .

$6 \times 6 = 36$

$-6 \times -6 = 36$

or

$y=\pm6$ is the correct answer.

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Question

Solve by finding the square root:

$\sqrt{49}*7$

Answer

First, find the square root:

$\sqrt{49}=7$

Then, solve:

Answer:

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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

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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