Question

Find the value of $\sqrt{0.01+\sqrt{0.0064}}$

Answer 1

Hint: To solve this question, we will first try to eliminate the square root using square, as $\sqrt{{{x}^{2}}}=x$. Squaring the square root gives us $0.01+\sqrt{0.0064}$. Now, we will assume ${{y}^{2}}$ as $0.01+\sqrt{0.0064}$. Taking the square of term $\sqrt{0.0064}$ we will arrive at 4 possible values of y. We will eliminate one which is not possible to get a result.

Complete step by step answer:
Let us assume, value of the given expression as y.
\[y=\sqrt{0.01+\sqrt{0.0064}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Now, we will try to eliminate the square root. So, squaring both sides, we get:
\[\begin{align}
  & {{y}^{2}}={{\left( \sqrt{0.01+\sqrt{0.0064}} \right)}^{2}} \\
 & {{y}^{2}}=0.01+\sqrt{0.0064}\text{ as }{{\left( \sqrt{x} \right)}^{2}}=x \\
\end{align}\]
Subtracting 0.01 both sides of above equation, we get:
\[{{y}^{2}}-0.01=\sqrt{0.0064}\]
Now, we observe that, again we have a square root left on RHS of above equation;
Therefore, squaring both sides we get:
\[{{\left( {{y}^{2}}-0.01 \right)}^{2}}={{\left( \sqrt{0.0064} \right)}^{2}}\]
Now, using the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab-{{b}^{2}}$ in LHS of above equation we get:
\[\begin{align}
  & {{\left( {{y}^{2}} \right)}^{2}}+{{\left( 0.01 \right)}^{2}}-2{{y}^{2}}\left( 0.01 \right)=0.0064\text{ as }{{\left( \sqrt{x} \right)}^{2}}=x \\
 & {{y}^{4}}+0.0001-0.02{{y}^{2}}=0.0064 \\
 & {{y}^{4}}-0.02{{y}^{2}}+0.0001-0.0064=0 \\
 & {{y}^{4}}-0.02{{y}^{2}}-0.0063=0 \\
\end{align}\]
Let ${{y}^{2}}=x$ using this substitution in above we get:
\[{{x}^{2}}-0.02x-0.0063=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
When we have an equation of the form $a{{x}^{2}}+bx+c=0$ then its roots are given as:
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Using $b=-0.02,c=-0.0063\text{ }\!\!\And\!\!\text{ }a=1$ in above equation (ii) we have
\[\begin{align}
  & x=\dfrac{+0.02\pm \sqrt{{{\left( 0.02 \right)}^{2}}-4\times 1\times \left( -0.0063 \right)}}{2} \\
 & x=\dfrac{0.02\pm \sqrt{0.0004+0.0252}}{2} \\
 & x=\dfrac{0.02\pm \sqrt{0.0256}}{2} \\
\end{align}\]
Now, we know that \[\sqrt{256}=\pm 16\Rightarrow \sqrt{0.0256}=0.16\text{ as }{{\left( \text{0}\text{.16} \right)}^{\text{2}}}=0.0256\]
Using this in above we get:
\[x=\dfrac{0.02\pm 0.16}{2}\]
Case I consider ‘+’ in between:
\[\begin{align}
  & x=\dfrac{0.02+0.16}{2} \\
 & x=\dfrac{0.18}{2}=0.09 \\
\end{align}\]
Then, ${{y}^{2}}=x=0.09\text{ and }{{y}^{2}}=0.09$
Taking square root both sides
\[\begin{align}
  & y=+\sqrt{0.09} \\
 & y=+0.3\text{ as }{{\left( 0.3 \right)}^{2}}=0.09 \\
\end{align}\]
Case II consider '-' in between:
\[\begin{align}
  & x=\dfrac{0.02-0.16}{2} \\
 & x=\dfrac{-0.14}{2}=-0.07 \\
\end{align}\]
Now, ${{y}^{2}}=x=-0.07\text{ and }{{y}^{2}}=-0.07$
Taking square root $y=\sqrt{-0.07}$
Now, because any negative inside the square root is not defined, so, value $y=\pm \sqrt{-0.07}$ is not defined.
Therefore, value of \[y=\sqrt{0.01+\sqrt{0.0064}}=+0.3\]

So, the correct answer is “Option A”.

Note: Another method to solve this question is
\[y=\sqrt{0.01+\sqrt{0.0064}}\]
Now, using $\sqrt{0.0064}=+0.08$ as ${{\left( 0.08 \right)}^{2}}=0.0064$
Using $\sqrt{0.0064}=0.08$ in above we get:
\[\begin{align}
  & y=\sqrt{0.01+0.08} \\
 & y=\sqrt{0.09} \\
\end{align}\]
Now, as value of $\sqrt{0.09}=+0.3$ we have $y=+0.3$ is answer here.