Question

Find the value of $k$ in $\sqrt{\sqrt{50}+\sqrt{48}}=k\left( \sqrt{3}+\sqrt{2} \right)$\[\]
A.${{2}^{\dfrac{1}{2}}}$\[\]
B. $2$\[\]
C. ${{2}^{\dfrac{1}{4}}}$\[\]
D. ${{2}^{\dfrac{1}{8}}}$\[\]

Answer 1

Hint: Square both side of the given equation. Use algebraic identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and do the prime factorization and use the formula $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ for the square root of left hand side. You will get a quadratic equation in $k$. Solve it to find the value of $k$.\[\]

Complete step-by-step solution:
The given expression is,
\[\sqrt{\sqrt{50}+\sqrt{48}}=k\left( \sqrt{3}+\sqrt{2} \right)\]
Let us first square both side of above equation we have,
\[\sqrt{50}+\sqrt{48}={{k}^{2}}{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}\]
Now we shall use the algebraic identity of square of sum of two numbers that is ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ in right hand side of the equation
\[\Rightarrow \sqrt{50}+\sqrt{48}={{k}^{2}}\left( 3+2+2\times \sqrt{2}\times \sqrt{3} \right)\]
We prime factorize the numbers on the left hand side and use the exponent identify $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ where $a, b$ are non-negative real numbers. We have $\sqrt{50}=\sqrt{2\times 5\times 5}=\sqrt{2}\times \sqrt{5\times 5}=5\sqrt{2}$ and $ sqrt{48}=\sqrt{2\times 2\times 2\times 2\times 3}=\sqrt{3}\times \sqrt{2\times 2\times 2\times 2}=4\sqrt{3}=2\sqrt{2}\sqrt{2}\sqrt{3}$. We can use $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ on the right side as $\sqrt{2}\times \sqrt{3}=\sqrt{2\times 3}=\sqrt{6}$. Putting all these values we have,
\[\begin{align}
  & \Rightarrow 5\sqrt{2}+4\sqrt{3}={{k}^{2}}\left( 5+2\sqrt{6} \right) \\
 & \Rightarrow 5\sqrt{2}+2\sqrt{2}\sqrt{2}\sqrt{3}={{k}^{2}}\left( 5+2\sqrt{6} \right) \\
\end{align}\]
Let us take common $\sqrt{2}$ on the left-hand side of the equation.
\[\Rightarrow \sqrt{2}\left( 5+2\sqrt{6} \right)={{k}^{2}}\left( 5+2\sqrt{6} \right)\]
Let us divide $5+2\sqrt{6}$ both side of the equation and get,
\[\Rightarrow \sqrt{2}={{k}^{2}}\]
The above equation is a quadratic equation which must have two roots. Let’s try to find them with linear polynomial factorization. We know the identity ${{a}^{m+n}}={{a}^{m}}+{{a}^{n}}$ where $a$ cannot be zero. Using it we have,
\[\begin{align}
  & \Rightarrow \sqrt{2}={{k}^{2}} \\
 & \Rightarrow {{k}^{2}}={{2}^{\dfrac{1}{2}}}={{2}^{\dfrac{1}{4}+\dfrac{1}{4}}}={{2}^{\dfrac{1}{4}}}\times {{2}^{\dfrac{1}{4}}} \\
 & \Rightarrow {{k}^{2}}-{{2}^{\dfrac{1}{4}}}\times {{2}^{\dfrac{1}{4}}}=0 \\
 & \Rightarrow \left( k-{{2}^{\dfrac{1}{4}}} \right)\left( k-{{2}^{\dfrac{1}{4}}} \right)=0 \\
\end{align}\]
So the roots of the quadratic equation $\sqrt{2}={{k}^{2}}$ has two roots and both of them are equal which are ${{2}^{\dfrac{1}{4}}}$. So the correct option is C.

Note: It is important to remember that the square is always positive because square root function inputs a non-negative real number and outputs a non-negative real number. While in case of solving an equation like ${{x}^{2}}=4$, we have to find the roots or zeroes of the equation .We do it by factorization or the quadratic formula of roots $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where $a,b,c$ are real numbers involved in the equation $a{{x}^{2}}+bx+c=0,a\ne 0$.