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Algebra II : Square Roots

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Example Question #21 : Square Roots

Evaluate: $\sqrt{0}-\sqrt1-\sqrt9-\sqrt{16}$ $\displaystyle \sqrt{0}-\sqrt1-\sqrt9-\sqrt{16}$

Possible Answers:

$\textup{Undefined.}$ $\displaystyle \textup{Undefined.}$

$\displaystyle 7$

$\displaystyle -7$

$\displaystyle -3$

$\displaystyle -8$

Correct answer:

$\displaystyle -8$

Explanation:

Evaluate each square root. The square root of a number evaluates into a number which multiplies by itself to achieve the number in the square root.

$\sqrt0 =0$ $\displaystyle \sqrt0 =0$

$\sqrt{1} = 1$ $\displaystyle \sqrt{1} = 1$

$\sqrt{9} =3$ $\displaystyle \sqrt{9} =3$

$\sqrt{16}=4$ $\displaystyle \sqrt{16}=4$

Substitute the terms back into the expression.

$\sqrt{0}-\sqrt1-\sqrt9-\sqrt{16} = 0-1-3-4 = -8$ $\displaystyle \sqrt{0}-\sqrt1-\sqrt9-\sqrt{16} = 0-1-3-4 = -8$

The answer is: $\displaystyle -8$

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Example Question #21 : Radicals

Solve: $\sqrt{16}+\sqrt{196}-\sqrt{36}$ $\displaystyle \sqrt{16}+\sqrt{196}-\sqrt{36}$

Possible Answers:

$\displaystyle 18$

$\displaystyle -18$

$\displaystyle -12$

$\displaystyle 12$

$\displaystyle 16$

Correct answer:

$\displaystyle 12$

Explanation:

Solve each radical. The square root determines a number that multiples by itself to equal the number inside the square root.

$\sqrt{16}=4$ $\displaystyle \sqrt{16}=4$

$\sqrt{196}=14$ $\displaystyle \sqrt{196}=14$

$\sqrt{36}=6$ $\displaystyle \sqrt{36}=6$

Rewrite the expression.

$\sqrt{16}+\sqrt{196}-\sqrt{36} = 4+14-6$ $\displaystyle \sqrt{16}+\sqrt{196}-\sqrt{36} = 4+14-6$

The answer is: $\displaystyle 12$

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Example Question #21 : Square Roots

True or false: $\sqrt{52}$ $\displaystyle \sqrt{52}$ is a radical expression in simplest form.

Possible Answers:

True

False

Correct answer:

False

Explanation:

A radical expression which is the $\displaystyle n$th root of a constant is in simplest form if and only if, when the radicand is expressed as the product of prime factors, no factor appears $\displaystyle n$ or more times. Since $\sqrt{52}$ $\displaystyle \sqrt{52}$ is a square, or second, root, find the prime factorization of 52, and determine whether any prime factor appears two or more times.

52 can be broken down as

$52 = 2 \times 26$ $\displaystyle 52 = 2 \times 26$

and further as

$52 = 2 \times 2 \times 13$ $\displaystyle 52 = 2 \times 2 \times 13$

The factor 2 appears twice, so $\sqrt{52}$ $\displaystyle \sqrt{52}$ is not in simplest form.

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Algebra II : Square Roots

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #21 : Square Roots

Example Question #21 : Radicals

Example Question #21 : Square Roots

All Algebra II Resources

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