
How do you simplify $4\sqrt {\dfrac{{81}}{{16}}} $.
Answer
456.6k+ views
Hint: We will first write the numerator and denominator separately in the form of squares. Then cut it with square root and multiply with 4 and thus we have the answer.
Complete step-by-step answer:
We need to simplify $4\sqrt {\dfrac{{81}}{{16}}} $. ………………(1)
We know that ${4^2} = 4 \times 4 = 16$ and ${9^2} = 9 \times 9 = 81$.
Since, we have $\sqrt {\dfrac{{81}}{{16}}} $ in the given expression which we require to simplify.
We can write this expression as: $\sqrt {\dfrac{{81}}{{16}}} = \sqrt {\dfrac{{{9^2}}}{{{4^2}}}} $ ……………..(2)
Now, we will make use of the fact that: $\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}$
On replacing a by 9 and b by 4, we get the following expression:-
$ \Rightarrow \dfrac{{{9^2}}}{{{4^2}}} = {\left( {\dfrac{9}{4}} \right)^2}$
Putting the above expression in the equation number (2), we will then obtain the following equation:-
\[ \Rightarrow \sqrt {\dfrac{{81}}{{16}}} = \sqrt {{{\left( {\dfrac{9}{4}} \right)}^2}} \] …………………..(3)
We also know that $\sqrt {{a^2}} = a$
On replacing a by $\dfrac{9}{4}$ in the above equation, we will then get the following equation:-
$ \Rightarrow \sqrt {{{\left( {\dfrac{9}{4}} \right)}^2}} = \dfrac{9}{4}$
Putting the above equation in the equation number (3), we will then obtain the following expression:-
\[ \Rightarrow \sqrt {\dfrac{{81}}{{16}}} = \dfrac{9}{4}\]
Putting the expression above in equation number (1), we will then obtain the following expression:-
$ \Rightarrow 4\sqrt {\dfrac{{81}}{{16}}} = 4 \times \dfrac{9}{4}$
Cutting off 4 from right hand side in above expression to get the following expression with us:-
$ \Rightarrow 4\sqrt {\dfrac{{81}}{{16}}} = 9$
Hence, we have $4\sqrt {\dfrac{{81}}{{16}}} $ simplified as 9.
Note:
The students must note that when we cut off 4 in the last third step it is because we know that 4 is not equal to 0.
The students must also note that when we open up square root, we have two possibilities, either a > 0 or a < 0. In general if nothing is given to us, we generally assume that a > 0 and thus we have the required answer as 9, otherwise it could have been -9 as well.
The students must note that the small things we use in the solution, sometimes we forget that they have a concept behind them like: $\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}$ and $\sqrt {{a^2}} = a$.
Complete step-by-step answer:
We need to simplify $4\sqrt {\dfrac{{81}}{{16}}} $. ………………(1)
We know that ${4^2} = 4 \times 4 = 16$ and ${9^2} = 9 \times 9 = 81$.
Since, we have $\sqrt {\dfrac{{81}}{{16}}} $ in the given expression which we require to simplify.
We can write this expression as: $\sqrt {\dfrac{{81}}{{16}}} = \sqrt {\dfrac{{{9^2}}}{{{4^2}}}} $ ……………..(2)
Now, we will make use of the fact that: $\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}$
On replacing a by 9 and b by 4, we get the following expression:-
$ \Rightarrow \dfrac{{{9^2}}}{{{4^2}}} = {\left( {\dfrac{9}{4}} \right)^2}$
Putting the above expression in the equation number (2), we will then obtain the following equation:-
\[ \Rightarrow \sqrt {\dfrac{{81}}{{16}}} = \sqrt {{{\left( {\dfrac{9}{4}} \right)}^2}} \] …………………..(3)
We also know that $\sqrt {{a^2}} = a$
On replacing a by $\dfrac{9}{4}$ in the above equation, we will then get the following equation:-
$ \Rightarrow \sqrt {{{\left( {\dfrac{9}{4}} \right)}^2}} = \dfrac{9}{4}$
Putting the above equation in the equation number (3), we will then obtain the following expression:-
\[ \Rightarrow \sqrt {\dfrac{{81}}{{16}}} = \dfrac{9}{4}\]
Putting the expression above in equation number (1), we will then obtain the following expression:-
$ \Rightarrow 4\sqrt {\dfrac{{81}}{{16}}} = 4 \times \dfrac{9}{4}$
Cutting off 4 from right hand side in above expression to get the following expression with us:-
$ \Rightarrow 4\sqrt {\dfrac{{81}}{{16}}} = 9$
Hence, we have $4\sqrt {\dfrac{{81}}{{16}}} $ simplified as 9.
Note:
The students must note that when we cut off 4 in the last third step it is because we know that 4 is not equal to 0.
The students must also note that when we open up square root, we have two possibilities, either a > 0 or a < 0. In general if nothing is given to us, we generally assume that a > 0 and thus we have the required answer as 9, otherwise it could have been -9 as well.
The students must note that the small things we use in the solution, sometimes we forget that they have a concept behind them like: $\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}$ and $\sqrt {{a^2}} = a$.
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