Question

Find the value of the positive square root of \[\sqrt{32}-\sqrt{24}.\]
\[\left( a \right)\sqrt[4]{2}\left( \sqrt{3}-1 \right)\]
\[\left( b \right)\sqrt[4]{2}\left( \sqrt{2}-1 \right)\]
\[\left( c \right)\sqrt[4]{2}\left( \sqrt{2}+1 \right)\]
\[\left( d \right)\sqrt[4]{2}\left( \sqrt{2}-\sqrt{3} \right)\]

Answer 1

Hint:To solve this question, we will first assume the variables for \[\sqrt{32}\] and \[\sqrt{24}\] separately and then make them in their simplest form by taking the LCM of 32 and 24 and converting the square root in the simplest form. After this process, we will subtract both what is obtained to get the result.

Complete step by step answer:
Let us consider \[\sqrt{32}-\sqrt{24}.\] Let us assume \[a=\sqrt{32}\] and \[b=\sqrt{24}.\]
Let us simplify ‘a’ first. The LCM of 32 is given as,
\[\begin{align}
  & 2\left| \!{\underline {\,
  32 \,}} \right. \\
 & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & \text{ 1} \\
\end{align}\]
\[\Rightarrow 32=2\times 2\times 2\times 2\times 2\]
\[\Rightarrow 32={{2}^{4}}\times 2\]
\[\Rightarrow \sqrt{32}=\sqrt{{{2}^{4}}\times 2}\]
\[\Rightarrow \sqrt{32}={{2}^{2}}\sqrt{2}\]
\[\Rightarrow \sqrt{32}=4\sqrt{2}\]
Therefore, we have got the value of a as
\[\Rightarrow a=\sqrt{32}=4\sqrt{2}.....\left( i \right)\]
Now, let us simplify ‘b’. The LCM of 24 is given as,
\[\begin{align}
  & 2\left| \!{\underline {\,
  24 \,}} \right. \\
 & 2\left| \!{\underline {\,
  12 \,}} \right. \\
 & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \text{ 1} \\
\end{align}\]
\[\Rightarrow 24=2\times 2\times 2\times 3\]
\[\Rightarrow 24={{2}^{2}}\times 2\times 3\]
\[\Rightarrow 24={{2}^{2}}\times 6\]
\[b=\sqrt{24}=\sqrt{{{2}^{2}}\times 6}\]
Therefore, we have got the value of b as
\[b=\sqrt{24}=2\sqrt{6}\]
Now because we need \[\sqrt{2}\] as it is obtained in equation (i). So, we have,
\[24=2\times 2\times 6\]
\[\Rightarrow \sqrt{24}=\sqrt{2}\sqrt{12}\]
So,
\[\Rightarrow b=\sqrt{24}=\sqrt{2}\sqrt{12}....\left( ii \right)\]
Now, compute a – b from equations (i) and (ii). We get,
\[\Rightarrow a-b=\sqrt{32}-\sqrt{24}=4\sqrt{2}-\sqrt{2}\sqrt{12}\]
\[\Rightarrow a-b=\sqrt{2}\left( 4-\sqrt{12} \right)\]
And, we know that \[\sqrt{12}=2\sqrt{3}\]
\[\Rightarrow \sqrt{32}-\sqrt{24}=\sqrt{2}\left( 4-2\sqrt{3} \right)\]
Finally, we have to compute the square root of \[\sqrt{32}-\sqrt{24}.\]
\[\Rightarrow \sqrt{\sqrt{32}\sqrt{24}}=\sqrt{\left( \sqrt{2} \right)}\sqrt{4-2\sqrt{3}}\]
\[\Rightarrow \sqrt{\sqrt{32}\sqrt{24}}={{\left( 2 \right)}^{\dfrac{1}{4}}}\sqrt{4-2\sqrt{3}}\]
Now, we will compute \[{{\left( \sqrt{3}-1 \right)}^{2}}\] using \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.\]
\[\Rightarrow {{\left( \sqrt{3}-1 \right)}^{2}}={{\left( \sqrt{3} \right)}^{2}}+1-2\sqrt{3}\]
\[\Rightarrow {{\left( \sqrt{3}-1 \right)}^{2}}=3+1-2\sqrt{3}\]
\[\Rightarrow {{\left( \sqrt{3}-1 \right)}^{2}}=4-2\sqrt{3}\]
\[\Rightarrow \sqrt{\sqrt{32}-\sqrt{24}}={{2}^{\dfrac{1}{4}}}\sqrt{{{\left( \sqrt{3}-1 \right)}^{2}}}\]
\[\Rightarrow \sqrt{\sqrt{32}-\sqrt{24}}={{2}^{\dfrac{1}{4}}}\left( \sqrt{3}-1 \right)\]
\[\Rightarrow \sqrt{\sqrt{32}-\sqrt{24}}=\sqrt[4]{2}\left( \sqrt{3}-1 \right)\]
Hence, option (a) is the right answer.

Note:
 Students need to note that the step where b is written as \[\sqrt{2}\times \sqrt{12}\] in place of \[2\sqrt{6}\] is necessary as we need some element common on both \[\sqrt{32}\] and \[\sqrt{24}\] so as to simplify the equation before obtaining the square root of them. \[2\sqrt{6}=b\] is also correct but it won’t give the longer result. The answer anyway would be the same.