Question

The approximate value of square root of 25.2 is

• A. 5.01
• B. 5.02
• C. 5.03
• D. 5.04

Answer 1

The correct option is B 5.02

Let $f\left(x\right)=\sqrt{x}$
Now, $f(x+\delta x)-f\left(x\right)=f^{\prime }\left(x\right).\delta x=\frac{\delta x}{2\sqrt{x}}$
We may write, 25.2 = 25+0.2
Taking x = 25 and $\delta x=0.2$ we have
$f\left(25.2\right)-f\left(25\right)=\frac{0.2}{2\sqrt{x}}$
We may write, 25.2 = 25+0.2
Taking x = 25 and $\delta x=0.2,$ we have
$f\left(25.2\right)-f\left(25\right)=\frac{0.2}{2\sqrt{25}}=0.02\therefore f\left(25.2\right)=f\left(25\right)+0.02=\sqrt{25}+0.02=5.02\Rightarrow \sqrt{\left(25.2\right)}=5.02$