Question

Simplify: StartRoot 16 r Superscript 6 Baseline EndRoot

4r2
4r3
8r2
8r3

Answer 1

Answer:

8r2 is my final answer because

Answer 2

Question: Simplify: $\sqrt{16r^6}$

(a) $4r^2$           (b) $4r^3$              (c) $8r^2$                (d) $8r^3$

Answer:

Option (b) is correct.

Step-by-step explanation:

We know that,

$\sqrt{x^2} =x$, for all positive integer $x$.

Consider the given root as follows:

$\sqrt{16r^6}$ . . . . . (1)

(a) $4r^2$

Let $x=4r^2$. Then,

Squaring both the sides as follows:

⇒ $x^2=(4r^2)^2$

⇒ $x^2=16r^4$ (Since $(a^m)^n=a^{mn}$)

Take square root both the sides.

⇒ $\sqrt{x^2}=\sqrt{16r^4}$

⇒ $x=\sqrt{16r^4}$

From (1), we have

$\sqrt{16r^6}\neq \sqrt{16r^4}$

Thus, option (a) is incorrect.

(b) $4r^3$

Let $x=4r^3$. Then,

Squaring both the sides as follows:

⇒ $x^2=(4r^3)^2$

⇒ $x^2=16r^6$ (Since $(a^m)^n=a^{mn}$)

Take square root both the sides.

⇒ $\sqrt{x^2}=\sqrt{16r^6}$

⇒ $x=\sqrt{16r^6}$

From (1), we have

$\sqrt{16r^6}= \sqrt{16r^6}$

Thus, option (b) is correct.

(c) $8r^2$

Let $x=8r^2$. Then,

Squaring both the sides as follows:

⇒ $x^2=(8r^2)^2$

⇒ $x^2=64r^4$ (Since $(a^m)^n=a^{mn}$)

Take square root both the sides.

⇒ $\sqrt{x^2}=\sqrt{64r^4}$

⇒ $x=\sqrt{64r^4}$

From (1), we have

$\sqrt{16r^6}\neq \sqrt{64r^4}$

Thus, option (c) is incorrect.

(d) $8r^3$

Let $x=8r^3$. Then,

Squaring both the sides as follows:

⇒ $x^2=(8r^3)^2$

⇒ $x^2=64r^6$ (Since $(a^m)^n=a^{mn}$)

Take square root both the sides.

⇒ $\sqrt{x^2}=\sqrt{64r^6}$

⇒ $x=\sqrt{64r^6}$

From (1), we have

$\sqrt{16r^6}\neq \sqrt{64r^6}$

Thus, option (d) is incorrect.

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