Question

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

Answer 1

Hint: Consider the square root of the above given number to be the sum of square roots of two numbers $x$ and $y$. Then square both the sides. Then you can compare the corresponding terms to find the values of $x$ and $y$ to solve the question.

Complete Step By Step Solution:
We have to find the square root of $10 + \sqrt {91} .$
Let the square root of $10 + \sqrt {91} = \sqrt x + \sqrt y $.
$ \Rightarrow \sqrt {10 + \sqrt {91} } = \sqrt x + \sqrt y $
On squaring on both sides, we get
\[{\left( {\sqrt {10 + \sqrt {91} } } \right)^2} = {\left( {\sqrt x + \sqrt y } \right)^2}\]
By using the formula, ${(a + b)^2} = {a^2} + {b^2} + 2ab$, we can expand the above equation as
$10 + \sqrt {91} = x + y + 2\sqrt {xy} .$
Now, for the equation to be true, the square root part in the LHS should be equal to the square root part of the RHS. And the remaining part of the LHS should be equal to the remaining part of the RHS. Therefore, by comparing both the sides, we can write
$x + y = 10$ . . . . . (1)
$2\sqrt {xy} = \sqrt {91} $ . . . . . (2)
By squaring equation (2), we get
${\left( {2\sqrt {xy} } \right)^2} = {\left( {\sqrt {91} } \right)^2}$
$ \Rightarrow 4xy = 91$
$ \Rightarrow xy = \dfrac{{91}}{4}$
$ \Rightarrow x = \dfrac{{91}}{{4y}}$
Put the value of $x$ in equation (1) to get the value of $y$
So, $x + y = 10$
$ \Rightarrow \dfrac{{91}}{{4y}} + y = 10$
Multiply both the sides by $y$
$ \Rightarrow 91 + 4{y^2} = 40y$
By rearranging it, we get
$ \Rightarrow 4{y^2} - 40y + 91 = 0$
This is a quadratic equation. We will solve it by factoring.
$ \Rightarrow 4{y^2} - 26y - 14y + 91 = 0$
$ \Rightarrow 2y(2y - 13) - 7(2y - 13) = 0$
By taking common term out, we get
$(2y - 13)(2y - 7) = 0$
This is true, if
$y = \dfrac{{13}}{2}$ or $y = \dfrac{7}{2}$
Put the value of$y$in equation (1) to find the value of $x$.
$x + y = 10$ . . . . (1)
When, $y = \dfrac{{13}}{2}$
We get
$x + \dfrac{{13}}{2} = 10$
$ \Rightarrow x = 10 - \dfrac{{13}}{2}$
\[ \Rightarrow x = \dfrac{7}{2}\]
When, $y = \dfrac{7}{2}$
We get
$x + \dfrac{7}{2} = 10$
$ \Rightarrow x = 10 - \dfrac{7}{2}$
$ \Rightarrow x = \dfrac{{13}}{2}$

Therefore, the square root of $10 + \sqrt {91} = \pm \left( {\sqrt {\dfrac{7}{2}} + \sqrt {\dfrac{{13}}{2}} } \right)$

Note:
Writing the square root as a sum of two variables is easy as we can easily form two equations by comparing the corresponding terms. We could have taken the square root to be one variable as well. But in that case, we would have to square the terms, two times, to remove two square root signs. By doing that, we would have got a bi-quadratic equation in one variable. Which could have been comparatively difficult to solve.