• Jan 27th 2009, 08:34 AM
mj.alawami

Q2) ³√ 54x^6

(all the integers are inside the square root )
Thank you,
• Jan 27th 2009, 08:39 AM
earboth
Quote:

Originally Posted by mj.alawami

Q2) ³√ 54x^6

(all the integers are inside the square root )
Thank you,

$\sqrt{48 x^2} = \sqrt{16x^2 \cdot 3} = \sqrt{(4x)^2 \cdot 3}$ ....... I'll leave the rest for you

$\sqrt[3]{54x^6} = \sqrt[3]{27x^6 \cdot 2} = \sqrt[3]{(3x^2)^3 \cdot 2}$ ....... I'll leave the rest for you
• Jan 27th 2009, 08:47 AM
mj.alawami
Is this equation correct

http://regentsprep.org/Regents/mathb/3B1/RatPo5.gif

Why can't it be =x^3/1 since there is no value before the square root

Thank you again
• Jan 27th 2009, 08:56 AM
earboth
Quote:

Originally Posted by mj.alawami
Is this equation correct

http://regentsprep.org/Regents/mathb/3B1/RatPo5.gif

Why can't it be =x^3/1 since there is no value before the square root

1. The equation is true.

2. $\sqrt{\ \ \ }$ is the short form of $\sqrt[2]{\ \ \ }$
• Jan 27th 2009, 08:59 AM
mj.alawami
Quote:

Originally Posted by earboth
1. The equation is true.

2. $\sqrt{\ \ \ }$ is the short form of $\sqrt[2]{\ \ \ }$

One final thing can you please convert the equations you sloved using fractions (So it will be easier for me to understand )

Thank you very much,I really appreciate it
• Jan 27th 2009, 09:10 AM
earboth
Quote:

Originally Posted by mj.alawami
One final thing can you please convert the equations you sloved using fractions (So it will be easier for me to understand )

Thank you very much,I really appreciate it

I'm not sure what you mean. There aren't any fractions involved. You only can convert the root sign into a rational exponent:

$
\sqrt{48 x^2} = \sqrt{16x^2 \cdot 3} = \sqrt{(4x)^2 \cdot 3} = \left((4x)^2 \cdot 3 \right)^{\frac12}
$

To change a cube root into a rational exponent use $\frac13$