QUESTION

# Find the square root of the following number : $19 + 8\sqrt 3$$A.{\text{ }}14 + \sqrt 3 \\ B.{\text{ }}4 + \sqrt 2 \\ C.{\text{ }}4 + \sqrt 3 \\ D.{\text{ }}2 + \sqrt 3 \\$

Hint: We had to assume that the square root of $19 + 8\sqrt 3$ is of the form $\sqrt a + \sqrt b$. And then we had to square both sides of the equation and compare to get the value of a and b.

As we know that the surd is an expression that includes a square root, cube root or other root symbol. Surds are used to write irrational numbers precisely – because the decimals of irrational numbers do not terminate or recur, they cannot be written exactly in decimal form.
So, we can see that the given number is the surd.
And, we cannot find the square root of surds by the prime factorization method.
So, as we know that the square root of the surd is also a surd or in other words, we can say that the square root of the form $\sqrt a + \sqrt b$.
So, let the square root of the given number be $\sqrt x + \sqrt y$.
So, $\sqrt {19 + 8\sqrt 3 } = \sqrt x + \sqrt y$ --- (1)
Now squaring both sides of the above equation to find the value of x and y. We get,
$19 + 8\sqrt 3$ = ${\left( {\sqrt x + \sqrt y } \right)^2}$ ---(2)
Now as we know the identity if a and b are two numbers then ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. So, applying this identity in equation 2. We get,
$19 + 8\sqrt 3 = {\left( {\sqrt x } \right)^2} + {\left( {\sqrt y } \right)^2} + 2\sqrt x \sqrt y$
$19 + 8\sqrt 3 = x + y + 2\sqrt {xy}$
Now if LHS and RHS of the above equation are equal then its root part and other part must be equal. So,
x + y = 19 ---(3)
And, $8\sqrt 3 = 2\sqrt {xy}$
Or, $\sqrt {xy} = 4\sqrt 3$ --- (4)
Now squaring equation 4. We get,
xy = 48 --- (5)
Now to find the value of x and y. We had to solve equation 3 and 5.
On dividing both sides of the equation 5 by y. We get,
$x = \dfrac{{48}}{y}$ ----(6)
Now putting the value of x in equation 3. We get,
$\dfrac{{48}}{y} + y = 19$
Taking LCM on LHS of the above equation. We get
$\dfrac{{48 + {y^2}}}{y} = 19$
Now cross multiplying both sides of the above equation. We get,
$48 + {y^2} = 19y$
${y^2} - 19y + 48 = 0$
Now we had to solve the above quadratic equation of y to get the required value of y.
So, splitting the middle term of the above quadratic equation. We get,
${y^2} - 16y - 3y + 48 = 0$
$y\left( {y - 16} \right) - 3\left( {y - 16} \right) = 0$
Now taking factor (y – 16) common from the above equation. We get,
$\left( {y - 16} \right)\left( {y - 3} \right) = 0$
So, y = 16 or y = 3
Putting the value of y in equation 6. We get,
When y = 16
Then, $x = \dfrac{{48}}{y} = \dfrac{{48}}{{16}} = 3$
When y = 3
Then, $x = \dfrac{{48}}{y} = \dfrac{{48}}{3} = 16$
So, the values of x and y are x = 16, y = 3 or x = 3, y = 16.
Now as we know that the square root of $19 + 8\sqrt 3$ is $\sqrt x + \sqrt y$.
So, putting the value of x and y in $\sqrt x + \sqrt y$. We get,
$\sqrt x + \sqrt y = \sqrt {16} + \sqrt 3$
Now as we know that the square root of 16 is 4. So,
$\sqrt x + \sqrt y = 4 + \sqrt 3$
So, the square root of $19 + 8\sqrt 3$ is $4 + \sqrt 3$.
Hence, the correct option will be C.

Note: We should always remember that the square root or cube root of any surd should also be a surd. So, if we are given a surd and asked to find the square root or cube root of that surd then we should assume another surd (here $\sqrt x + \sqrt y$) as the square root of the given number and then we equate the square root of the given number by $\sqrt x + \sqrt y$ and after that we square both sides of the equation. After that we form two equations by comparing the root part and the perfect number of both sides of the equation. And then solve both of the equations by using a substitution method to get the required value of x and y. And then $\sqrt x + \sqrt y$ will be the square root of the given number.