Question

Find the square root of \[\sqrt{48}+\sqrt{45}\].

Answer 1

Hint: First look at the definition of prime factorization and do the prime factorization of both the numbers inside the root. Now add their square roots, try to make a perfect square. Then apply square root on both sides of the equation. The value you get would be the result required.

Complete step-by-step answer:
In number theory, prime factorization is the decomposition of a composite number into a product of few prime numbers which are smaller than the original number. This process is carried out by dividing with prime numbers and finding quotients. Thus writing numbers as \[prime\times quotient\] Repeat the process for the quotient till you get 1 as the quotient.
Prime factorization of 48, by dividing with 2, we can write:
\[48=24\times 2\]
By dividing term 24 with 2, we can write 48 as:
\[48=12\times 2\times 2\]
By dividing term 12 with 2, we can write 48 as:
\[48=6\times 2\times 2\times 2\]
By dividing term 6 with 2, we can write 48 as:
\[48=3\times 2\times 2\times 2\times 2\]
So, by applying square root on both sides, we get it as:
\[\sqrt{48}=\sqrt{3\times {{2}^{4}}}\]
By simplifying the above, we can write it as:
\[\sqrt{48}=4\sqrt{3}\ldots \ldots \ldots \ldots ..\left( \text{1} \right)\]
Prime factorization of 45, by dividing with 5, we can write:
\[45=5\times 9\]
As 9 can be written as $ {{3}^{2}} $ we get:
\[45=5\times {{3}^{2}}\]
By applying square root on both sides, we get it as:
\[\sqrt{45}=3\sqrt{5}\ldots \ldots \ldots \ldots ..\left( \text{2} \right)\]
By adding equating (1), (2) we get an equation written as:
\[\sqrt{48}+\sqrt{45}=4\sqrt{3}+3\sqrt{5}\]
By taking $ \sqrt{3} $ as common, we can write the equation as:
\[\sqrt{48}+\sqrt{45}=\sqrt{3}\left( 4+\sqrt{3}.\sqrt{5} \right)\]
By multiplying abs dividing with 2 on right hand side, we get:
\[\sqrt{48}+\sqrt{45}=\dfrac{\sqrt{3}}{2}\left( 8+2\sqrt{3}\sqrt{5} \right)=\dfrac{\sqrt{3}}{2}\left( 5+3+2\sqrt{3}\sqrt{5} \right)\]
As we know \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\] by substituting $ a=\sqrt{5\,}\,,\,\,b=\sqrt{3\,} $ we get:
\[\sqrt{\sqrt{48}+\sqrt{45}}=\sqrt{\dfrac{\sqrt{3}}{2}{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}}=\sqrt{\dfrac{\sqrt{3}}{2}}\left( \sqrt{5}+\sqrt{3} \right)\]
Note: Use of the prime factorization method is very important in this question. You must do till you get 1. Hence, in case of 45 we get 9 which is square of a prime. That’s the reason we stopped at this step. So, we can cancel squares with roots. At last turning the sum into terms of perfect square is an important step.