Which of the following expressions is equivalent to 5sqrt(98a^(9)b^(4)) ? Choose 1 answer: (A) 5a^(3)b^(2)sqrt98 (B) 10a^(4)b^(2)sqrt(7a) (c) 35a^(4)b^(2)sqrt(2a) (D) 245a^(3)b^(2)sqrt2

Factor Perfect Squares: First, let's simplify the expression inside the square root by factoring out perfect squares. 98 can be factored into 49 * 2, where 49 is a perfect square (7^2). a^9 can be factored into a^8 * a, where a^8 is a perfect square (a^4)^2. b^4 is already a perfect square (b^2)^2. So, we can rewrite the expression as 5sqrt(49 * 2 * a^8 * a * b^4).Take Square Roots: Next, we take the square root of the perfect squares. The square root of 49 is 7. The square root of a^8 is a^4. The square root of b^4 is b^2. So, we can take these outside the square root, giving us 5 * 7 * a^4 * b^2 * sqrt(2a).Multiply Constants: Now, we multiply the constants outside the square root. 5 * 7 equals 35. So, the expression simplifies to 35a^4b^2sqrt(2a).Compare Answer Choices: Finally, we compare our simplified expression to the answer choices. The expression 35a^4b^2sqrt(2a) matches with option (C).

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Which of the following expressions is equivalent to
5sqrt(98a^(9)b^(4)) ?
Choose 1 answer:
(A)
5a^(3)b^(2)sqrt98
(B)
10a^(4)b^(2)sqrt(7a)
(c)
35a^(4)b^(2)sqrt(2a)
(D)
245a^(3)b^(2)sqrt2

Which of the following expressions is equivalent to $5 \sqrt{98 a^{9} b^{4}}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $5 a^{3} b^{2} \sqrt{98}$ $\newline$ (B) $10 a^{4} b^{2} \sqrt{7 a}$ $\newline$ (C) $35 a^{4} b^{2} \sqrt{2 a}$ $\newline$ (D) $245 a^{3} b^{2} \sqrt{2}$

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Math Problems

Algebra 2

Simplify radical expressions involving fractions

Full solution

Q. Which of the following expressions is equivalent to

5 \sqrt{98 a^{9} b^{4}}

\newline

Choose

1

answer:

\newline

(A)

5 a^{3} b^{2} \sqrt{98}

\newline

(B)

10 a^{4} b^{2} \sqrt{7 a}

\newline

(C)

35 a^{4} b^{2} \sqrt{2 a}

\newline

(D)

245 a^{3} b^{2} \sqrt{2}

Factor Perfect Squares: First, let's simplify the expression inside the square root by factoring out perfect squares. $\newline$ $98$ can be factored into $49 \times 2$ , where $49$ is a perfect square $(7^2)$ . $\newline$ $a^9$ can be factored into $a^8 \times a$ , where $a^8$ is a perfect square $(a^4)^2$ . $\newline$ $b^4$ is already a perfect square $(b^2)^2$ . $\newline$ So, we can rewrite the expression as $49 \times 2$ $0$ .
Take Square Roots: Next, we take the square root of the perfect squares. $\newline$ The square root of $49$ is $7$ . $\newline$ The square root of $a^8$ is $a^4$ . $\newline$ The square root of $b^4$ is $b^2$ . $\newline$ So, we can take these outside the square root, giving us $5 \times 7 \times a^4 \times b^2 \times \sqrt{2a}$ .
Multiply Constants: Now, we multiply the constants outside the square root. $5 \times 7$ equals $35$ . So, the expression simplifies to $35a^4b^2\sqrt{2a}$ .
Compare Answer Choices: Finally, we compare our simplified expression to the answer choices. The expression $35a^4b^2\sqrt{2a}$ matches with option (C).