How do you simplify $\sqrt{0.36}$?
Answer
585k+ views
Hint:We try to form the indices formula for the value 2. This is a square root of $0.36$. We convert the decimal to fractions. The root of the whole fraction happens by finding the root of the denominator and numerator separately as \[\sqrt{\dfrac{p}{q}}=\dfrac{\sqrt{p}}{\sqrt{q}}\].
We find the root values and place to find the solution. We find the prime factorisation of 36 and 100.
Complete step by step solution:
We need to find the value of the algebraic form of $\sqrt{0.36}$. This is a square root form.
The given value is the form of indices. We are trying to find the root value of $0.36$.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get
\[{{a}^{\dfrac{1}{2}}}=\sqrt{a}\].
Converting decimal to fraction we get $0.36=\dfrac{36}{100}$.
We have to solve the answer of $\sqrt{\dfrac{36}{100}}$. We know
\[\sqrt{\dfrac{p}{q}}=\dfrac{\sqrt{p}}{\sqrt{q}}\].
This means to find the answer of $\sqrt{\dfrac{36}{100}}$, we need to find \[\sqrt{36}\] and
$\sqrt{100}$.
For \[\sqrt{36}=\sqrt{{{\left( 6 \right)}^{2}}}=6\] and for \[\sqrt{100}=\sqrt{{{\left( 10
\right)}^{2}}}=10\]
We now place the values and get
$\sqrt{\dfrac{36}{100}}=\dfrac{\sqrt{36}}{\sqrt{100}}=\dfrac{6}{10}=\dfrac{3}{5}$.
Now we convert the fraction into fraction and get $\dfrac{3}{5}=0.6$
The simplified value of the square root $\sqrt{0.36}$ is $0.6$.
Note: We can also use the variable from where we can take $x={{\left( 0.36 \right)}^{\dfrac{1}{2}}}$. But we need to remember that we can’t use the square on both sides of the equation $x={{\left( 0.36 \right)}^{\dfrac{1}{2}}}$ as in that case we are taking an extra value of negative as in $0.36$ as a root value. Then this linear equation becomes a quadratic equation.
We find the root values and place to find the solution. We find the prime factorisation of 36 and 100.
Complete step by step solution:
We need to find the value of the algebraic form of $\sqrt{0.36}$. This is a square root form.
The given value is the form of indices. We are trying to find the root value of $0.36$.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get
\[{{a}^{\dfrac{1}{2}}}=\sqrt{a}\].
Converting decimal to fraction we get $0.36=\dfrac{36}{100}$.
We have to solve the answer of $\sqrt{\dfrac{36}{100}}$. We know
\[\sqrt{\dfrac{p}{q}}=\dfrac{\sqrt{p}}{\sqrt{q}}\].
This means to find the answer of $\sqrt{\dfrac{36}{100}}$, we need to find \[\sqrt{36}\] and
$\sqrt{100}$.
For \[\sqrt{36}=\sqrt{{{\left( 6 \right)}^{2}}}=6\] and for \[\sqrt{100}=\sqrt{{{\left( 10
\right)}^{2}}}=10\]
We now place the values and get
$\sqrt{\dfrac{36}{100}}=\dfrac{\sqrt{36}}{\sqrt{100}}=\dfrac{6}{10}=\dfrac{3}{5}$.
Now we convert the fraction into fraction and get $\dfrac{3}{5}=0.6$
The simplified value of the square root $\sqrt{0.36}$ is $0.6$.
Note: We can also use the variable from where we can take $x={{\left( 0.36 \right)}^{\dfrac{1}{2}}}$. But we need to remember that we can’t use the square on both sides of the equation $x={{\left( 0.36 \right)}^{\dfrac{1}{2}}}$ as in that case we are taking an extra value of negative as in $0.36$ as a root value. Then this linear equation becomes a quadratic equation.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

10 examples of friction in our daily life

Name the Largest and the Smallest Cell in the Human Body ?

