Question

divide -39 x^4 by square root 13x^2​

Answer 1

Step-by-step explanation:

finding root square of 13 is by the formula of √x+y= √x±y/2×√x.

Answer 2

Answer:  $-3\sqrt{13} x^3$

Step-by-step explanation:

We have to divide $-39x^4$ by $\sqrt{13x^2}$

Therefore

$\dfrac{-39x^4}{\sqrt{13x^2} } = \dfrac{-39x^4}{\sqrt{13} \sqrt{x^2} } \\\\\\= \dfrac{-39x^4}{\sqrt{13}\times x}$

Now rationalize we get

$\dfrac{-39x^4}{\sqrt{13}x }\times \dfrac{\sqrt{13} }{\sqrt{13} } = \dfrac{-39\sqrt{13}x^{4} }{13x}$

Now as we know $a^m\div a^n = a^{m-n}$

$\dfrac{-39\sqrt{13}x^{4-1} }{13}= \dfrac{-39\sqrt{13}x^3 }{13}= -3\sqrt{13} x^3$

Hence, the required answer is $-3\sqrt{13} x^3$