
How do you find the square root $\sqrt {1.44} $?
Answer
542.1k+ views
Hint: Here we can proceed by writing the term which is $1.44$ as $\dfrac{{144}}{{100}}$ which means we need to find $\sqrt {\dfrac{{144}}{{100}}} $ and now we can write it in the form $\dfrac{{\sqrt {144} }}{{\sqrt {100} }}$ and then find their roots separately and put their values here to get the root of the decimal $1.44$.
Complete step by step solution:
Here we are given to find the square root of the term $1.44$ which means we need to find the number which when multiplied by itself gives us the value $1.44$.
So we can write $1.44$ as $\dfrac{{144}}{{100}}$.
Now we know that we need to find the square root of $\sqrt {\dfrac{{144}}{{100}}} $ and we must know that whenever we have the two square roots in the numerator and denominator we can find their individual roots and then divide in order to get the root of the required decimal term.
So here we can write $\sqrt {\dfrac{{144}}{{100}}} $$ = \dfrac{{\sqrt {144} }}{{\sqrt {100} }}$$ - - - (1)$.
Now we know that $\sqrt {100} = \sqrt {(10)(10)} = 10$ and also $\sqrt {144} = \sqrt {(2)(2)(2)(2)(3)(3)} = (2)(2)(3) = 12$.
So we can say that the square root of the denominator is $100$ as $10$ because when we multiply it twice we get the number at $100$.
The square root of the numerator is $144$ as $12$ because when it is multiplied twice we get the term as $144$.
Now we can easily substitute their values in equation (1) to get the value of the decimal number which is $1.44$.
So putting these values we get:
$\sqrt {\dfrac{{144}}{{100}}} $$ = \dfrac{{\sqrt {144} }}{{\sqrt {100} }}$$ = \dfrac{{12}}{{10}}$$ = 1.2$
Hence we get the square root of $1.44$ as $1.2$.
This can also be checked by multiplying the term $1.2$ with itself and we will see that we will get the resultant after multiplication as $1.44$.
Note:
Here in these types of problems, the student must be clear with the difference between the term square and the square root. The square root of the number is the number that needs to be multiplied with itself to get the number inside the root while the square of the number is just to multiply it twice and get the result. For example: Square of $3$ is $9$ and square root of $9{\text{ is 3}}$.
Complete step by step solution:
Here we are given to find the square root of the term $1.44$ which means we need to find the number which when multiplied by itself gives us the value $1.44$.
So we can write $1.44$ as $\dfrac{{144}}{{100}}$.
Now we know that we need to find the square root of $\sqrt {\dfrac{{144}}{{100}}} $ and we must know that whenever we have the two square roots in the numerator and denominator we can find their individual roots and then divide in order to get the root of the required decimal term.
So here we can write $\sqrt {\dfrac{{144}}{{100}}} $$ = \dfrac{{\sqrt {144} }}{{\sqrt {100} }}$$ - - - (1)$.
Now we know that $\sqrt {100} = \sqrt {(10)(10)} = 10$ and also $\sqrt {144} = \sqrt {(2)(2)(2)(2)(3)(3)} = (2)(2)(3) = 12$.
So we can say that the square root of the denominator is $100$ as $10$ because when we multiply it twice we get the number at $100$.
The square root of the numerator is $144$ as $12$ because when it is multiplied twice we get the term as $144$.
Now we can easily substitute their values in equation (1) to get the value of the decimal number which is $1.44$.
So putting these values we get:
$\sqrt {\dfrac{{144}}{{100}}} $$ = \dfrac{{\sqrt {144} }}{{\sqrt {100} }}$$ = \dfrac{{12}}{{10}}$$ = 1.2$
Hence we get the square root of $1.44$ as $1.2$.
This can also be checked by multiplying the term $1.2$ with itself and we will see that we will get the resultant after multiplication as $1.44$.
Note:
Here in these types of problems, the student must be clear with the difference between the term square and the square root. The square root of the number is the number that needs to be multiplied with itself to get the number inside the root while the square of the number is just to multiply it twice and get the result. For example: Square of $3$ is $9$ and square root of $9{\text{ is 3}}$.
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