Question

Find the square root of $10+\sqrt{91}$. \[\]

Answer 1

Hint: We assume $x$ and $y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{10+\sqrt{91}}$. We square left hand side using identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and then compare the rational and irrational parts to find $x$ and $y$ after solving a quadratic equation in either $x$ or $y$. \[\]

Complete step by step answer:
We are asked to find the square root of $10+\sqrt{91}$ which means the value of $\sqrt{10+\sqrt{91}}$. Let us assume $x$ and $y$ such that
\[\sqrt{x}+\sqrt{y}=\sqrt{10+\sqrt{91}}......\left( 1 \right)\]
. Let us square both side of the equations to have;
\[{{\left( \sqrt{x}+\sqrt{y} \right)}^{2}}={{\left( \sqrt{10+\sqrt{91}} \right)}^{2}}\]
Let us use the algebraic identity of whole square of two numbers ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ for $a=\sqrt{x},b=\sqrt{y}$ in the left hand side a of the equation to expand to have;
\[\Rightarrow x+y+2\sqrt{xy}=10+\sqrt{91}.....\left( 2 \right)\]
Let us compare the rational part of the above equation and have;
\[x+y=10......\left( 3 \right)\]
Let us compare the irrational part of the above equation(2) and have;
\[\begin{align}
  & 2\sqrt{xy}=\sqrt{91} \\
 & \Rightarrow \sqrt{xy}=\dfrac{\sqrt{91}}{2} \\
\end{align}\]
We square both sides of above equation to have;
\[\begin{align}
  & xy=\dfrac{91}{4} \\
 & \Rightarrow x=\dfrac{91}{4y} \\
\end{align}\]
We put the value of $x$in equation (3) to have;
\[\begin{align}
  & x+y=10 \\
 & \Rightarrow \dfrac{91}{4y}+y=10 \\
 & \Rightarrow \dfrac{91+4{{y}^{2}}}{4y}=10 \\
 & \Rightarrow 4{{y}^{2}}-40y+91=0 \\
\end{align}\]
We solve the above quadratic equation in $y$ by splitting the middle term method.
\[\begin{align}
  & \Rightarrow 4{{y}^{2}}-26y-14y+91=0 \\
 & \Rightarrow 2y\left( 2y-13 \right)-7\left( 2y-13 \right)=0 \\
 & \Rightarrow \left( 2y-13 \right)\left( 2y-7 \right)=0 \\
 & \Rightarrow 2y-13=0\text{ or }2y-7=0 \\
\end{align}\]
So we have two solutions for $y$ as
\[y=\dfrac{13}{2},y=\dfrac{7}{2}\]
We put the value of $y=\dfrac{13}{2}$ in equation (3) and find corresponding value of $x$ as;
\[\begin{align}
  & x+\dfrac{13}{2}=10 \\
 & \Rightarrow x=10-\dfrac{13}{2} \\
 & \Rightarrow x=\dfrac{7}{2} \\
\end{align}\]
We put the value of $y=\dfrac{7}{2}$ in equation (3) and find corresponding value of $x$ as;
\[\begin{align}
  & x+\dfrac{7}{2}=10 \\
 & \Rightarrow x=10-\dfrac{7}{2} \\
 & \Rightarrow x=\dfrac{13}{2} \\
\end{align}\]
We put the values of $x=\dfrac{7}{2},y=\dfrac{13}{2}$ or $x=\dfrac{13}{2},y=\dfrac{7}{2}$ in equation (1) find the required value of $\sqrt{10+\sqrt{91}}$same because of commutative nature of addition as;

\[\sqrt{10+\sqrt{91}}=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{13}{2}}\]

Note: We note that we should not assign a single variable to square like $x=10+\sqrt{91}$ because then we have to square twice and calculation will be difficult. The number $10+\sqrt{91}$ is surd and two surds $a+\sqrt{b},c+\sqrt{d}$ are going to be equal when their rational part are equal $a=b$ and irrational parts are equal $c=d$. We can find the factors for splitting the middle term from the prime factorization of $4\times 91$ or we can use a quadratic formula.