Question

The square root of $\frac{(0.75)^{3}}{1-(0.75)}+\left[0.75+(0.75)^{2}+1\right]$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Let $x=0.75$
According to the question,
$\frac{x^{3}}{1-x}+\left(x+x^{2}+1\right)$
$=\frac{x^{3}+(1-x)\left(1+x+x^{2}\right)}{1-x}$
$=\frac{x^{3}+1-x^{3}}{1-x}=\frac{1}{1-x}$
Now, put the value of $x$
$\frac{1}{1-0.75}=\frac{1}{0.25}$
$=\frac{100}{25}=4$
So, square root of the equation $=\sqrt{4}=2$