Question

Let $x=\sqrt{3-\sqrt{5}}$ and $y=\sqrt{3+\sqrt{5}}$. If the value of the expression $x-y+2{{x}^{2}}y+2x{{y}^{2}}-{{x}^{4}}y+x{{y}^{4}}$ can be expressed in the form $\sqrt{p}+\sqrt{q}$ where $p,q\in N$, then $\left( p+q \right)$ has the value equal to?

Answer 1

Hint: Assume the given expression as E. Now, multiply x and y, simplify using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. In the expression E take $2xy$ common from the terms $2{{x}^{2}}y+2x{{y}^{2}}$ and $-xy$ common from the terms $-{{x}^{4}}y+x{{y}^{4}}$. Use the algebraic identities ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to find the value of (x + y), ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to find the value of (x – y) and $\left( {{x}^{3}}-{{y}^{3}} \right)={{\left( x-y \right)}^{3}}+3xy\left( x-y \right)$ to simplify the expression. Substitute all the obtained values to get the answer.

Complete step by step solution:
Here we have been provided with values $x=\sqrt{3-\sqrt{5}}$ and $y=\sqrt{3+\sqrt{5}}$, we are asked to find the value of the expression $x-y+2{{x}^{2}}y+2x{{y}^{2}}-{{x}^{4}}y+x{{y}^{4}}$ in the form of $\sqrt{p}+\sqrt{q}$ then we have to find the value of $\left( p+q \right)$. Here p and q are natural numbers. Let us assume the expression as E, so we have,
$\begin{align}
  & \Rightarrow E=x-y+2{{x}^{2}}y+2x{{y}^{2}}-{{x}^{4}}y+x{{y}^{4}} \\
 & \Rightarrow E=\left( x-y \right)+2xy\left( x+y \right)-xy\left( {{x}^{3}}-{{y}^{3}} \right) \\
\end{align}$
Using the algebraic identity $\left( {{x}^{2}}-{{y}^{3}} \right)={{\left( x-y \right)}^{3}}+3xy\left( x-y \right)$ we get,
$\Rightarrow E=\left( x-y \right)+2xy\left( x+y \right)-xy\left[ {{\left( x-y \right)}^{3}}+3xy\left( x-y \right) \right]$ …….. (1)
Now let us find the values of $xy$, $\left( x+y \right)$ and $\left( x-y \right)$ using certain algebraic identities like $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ respectively.
(i) Multiplying the expressions of x and y we get,
$\begin{align}
  & \Rightarrow xy=\sqrt{3-\sqrt{5}}\times \sqrt{3+\sqrt{5}} \\
 & \Rightarrow xy=\sqrt{\left( 3-\sqrt{5} \right)\left( 3+\sqrt{5} \right)} \\
\end{align}$
Using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ we get,
$\begin{align}
  & \Rightarrow xy=\sqrt{{{\left( 3 \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}} \\
 & \Rightarrow xy=\sqrt{9-5} \\
 & \Rightarrow xy=\sqrt{4} \\
 & \Rightarrow xy=2 \\
\end{align}$
(ii) Using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and substituting x = a and y = b we get,
$\begin{align}
  & \Rightarrow {{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy \\
 & \Rightarrow {{\left( x+y \right)}^{2}}=3-\sqrt{5}+3+\sqrt{5}+4 \\
 & \Rightarrow {{\left( x+y \right)}^{2}}=10 \\
\end{align}$
Taking square root both the sides we get,
$\Rightarrow \left( x+y \right)=\sqrt{10}$
(iii) Using the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and substituting x = a and y = b we get,
$\begin{align}
  & \Rightarrow {{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy \\
 & \Rightarrow {{\left( x-y \right)}^{2}}=3-\sqrt{5}+3+\sqrt{5}-4 \\
 & \Rightarrow {{\left( x-y \right)}^{2}}=2 \\
\end{align}$
Taking square root both the sides and considering the negative value because we can clearly see that y is greater than x, so we get,
$\Rightarrow \left( x-y \right)=-\sqrt{2}$
Now, substituting the above obtained values of three expressions in equation (1) we get,
$\begin{align}
  & \Rightarrow E=-\sqrt{2}+2\times 2\times \sqrt{10}-2\left[ {{\left( -\sqrt{2} \right)}^{3}}+3\times 2\times \left( -\sqrt{2} \right) \right] \\
 & \Rightarrow E=4\sqrt{10}+15\sqrt{2} \\
\end{align}$
Taking the terms inside the square root and comparing with the form $\sqrt{p}+\sqrt{q}$ we get,
$\begin{align}
  & \Rightarrow E=\sqrt{16\times 10}+\sqrt{225\times 2} \\
 & \Rightarrow \sqrt{p}+\sqrt{q}=\sqrt{160}+\sqrt{450} \\
\end{align}$
On comparing we can assume p = 160 and q = 450, so adding p and q we get,
$\begin{align}
  & \Rightarrow p+q=160+450 \\
 & \therefore p+q=610 \\
\end{align}$
Hence, the value of the expression (p + q) is equal to 610.

Note: Note that you have to be careful while finding the value of the expression $\left( x-y \right)$ because if you will take its value equal to $\sqrt{2}$ instead of $-\sqrt{2}$ then you will get the expression of the form $\sqrt{p}-\sqrt{q}$ and therefore we will not be able to find the value of p + q. Remember all the algebraic identities and do not try to find the value of the expressions x and y by substituting the approx decimal value of $\sqrt{5}$.