Question

What is the square root of $\frac{121a^4{b}^6{c}^8}{196t^{14}{s}^6}$?

• A. $\frac{11}{14}\left|\frac{a^2{b}^3{c}^4}{t^7{s}^3}\right|$
• B. $\frac{11}{14}\frac{a^2{b}^3{c}^4}{t^7{s}^3}$
• C. $\frac{11}{14}\left|\frac{a^2{b}^8{c}^4}{t^7{s}^4}\right|$
• D. $\frac{1}{14}\left|\frac{a^2{b}^3{c}^4}{t^7{s}^3}\right|$

Answer 1

The correct option is A $\frac{11}{14}\left|\frac{a^2{b}^3{c}^4}{t^7{s}^3}\right|$

Given, $\frac{121a^4{b}^6{c}^8}{196t^{14}{s}^6}$

$\Rightarrow \sqrt{\frac{121a^4{b}^6{c}^8}{196t^{14}{s}^6}}$

$=$ $\sqrt{{\left(\frac{11a^2{b}^3{c}^4}{14t^7{s}^3}\right)}^2}$

So, the square root of $64(x-y{)}^4(y+x{)}^6(x-z{)}^{1}0=\frac{11}{14}\left|\frac{a^2{b}^3{c}^4}{t^7{s}^3}\right|$