Find the square root of the expression, $11+2\sqrt{30}$.
Answer
672.3k+ views
Hint: In order to find the solution of this question, we will consider the square root of $11+2\sqrt{30}$ as $a+\sqrt{b}$ because the square root of an irrational number is always irrational. Also, we should know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$. By using this we will compare the root of $11+2\sqrt{30}$ by it and we will get the answer.
Complete step-by-step answer:
In this question, we have been asked to find the square root of $11+2\sqrt{30}$. To solve this question, we will consider $a+\sqrt{b}$ as the square root of $11+2\sqrt{30}$ because the square root of an irrational number is always irrational. So, we can write,
${{\left( a+\sqrt{b} \right)}^{2}}=11+2\sqrt{30}$
Now, we know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$. So, for x = a and $y=\sqrt{b}$, we can say, ${{\left( a+\sqrt{b} \right)}^{2}}={{a}^{2}}+b+2a\sqrt{b}$. Therefore, we can write,
${{a}^{2}}+b+2a\sqrt{b}=11+2\sqrt{30}$
Now, we will compare the rational and irrational part of both sides of the equation. So, we get,
${{a}^{2}}+b=11$ and $2a\sqrt{b}=2\sqrt{30}$
$\Rightarrow {{a}^{2}}+b=11.........\left( i \right)$ and $a=\dfrac{\sqrt{30}}{\sqrt{b}}.........\left( ii \right)$
From equation (ii), we will put the value of a in equation (i). So, we get,
$\begin{align}
& {{\left( \dfrac{\sqrt{30}}{\sqrt{b}} \right)}^{2}}+b=11 \\
& \dfrac{30}{b}+b=11 \\
& \dfrac{30+{{b}^{2}}}{b}=11 \\
& 30+{{b}^{2}}=11b \\
& {{b}^{2}}-11b+30=0 \\
\end{align}$
And, by middle term spitting method, we can write,
$\begin{align}
& {{b}^{2}}-5b-6b+30=0 \\
& b\left( b-5 \right)-6\left( b-5 \right)=0 \\
& \left( b-5 \right)\left( b-6 \right)=0 \\
\end{align}$
$b-5=0$ or $b-6=0$
b = 5 or b = 6
Now, we will put both the values of b one by one in equation (ii). So, we get,
For b = 5, $a=\dfrac{\sqrt{30}}{\sqrt{5}}=\sqrt{6}$ and for b = 6, $a=\dfrac{\sqrt{30}}{\sqrt{6}}=\sqrt{5}$.
Therefore, for b = 5 and $a=\sqrt{6}$, we get $\left( a+\sqrt{b} \right)=\left( \sqrt{6}+\sqrt{5} \right)$ and, for b = 6 and $a=\sqrt{5}$, we get $\left( a+\sqrt{b} \right)=\left( \sqrt{5}+\sqrt{6} \right)$. In both the cases we got \[\left( a+\sqrt{b} \right)=\left( \sqrt{5}+\sqrt{6} \right)\]. Hence, we can say that the square root of $11+2\sqrt{30}$ is \[\left( \sqrt{5}+\sqrt{6} \right)\].
Note: We can also solve this question by hit and trial method, by writing $\sqrt{30}=\sqrt{2}\times \sqrt{15}=\sqrt{6}\times \sqrt{5}=1\times \sqrt{30}$ and then by choosing the correct possible pair, which might give us the correct answer for this small value. But if we get some bigger value and then apply the hit and trial method, it will waste a lot of time. So, it is better to solve this question using the conventional method.
Complete step-by-step answer:
In this question, we have been asked to find the square root of $11+2\sqrt{30}$. To solve this question, we will consider $a+\sqrt{b}$ as the square root of $11+2\sqrt{30}$ because the square root of an irrational number is always irrational. So, we can write,
${{\left( a+\sqrt{b} \right)}^{2}}=11+2\sqrt{30}$
Now, we know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$. So, for x = a and $y=\sqrt{b}$, we can say, ${{\left( a+\sqrt{b} \right)}^{2}}={{a}^{2}}+b+2a\sqrt{b}$. Therefore, we can write,
${{a}^{2}}+b+2a\sqrt{b}=11+2\sqrt{30}$
Now, we will compare the rational and irrational part of both sides of the equation. So, we get,
${{a}^{2}}+b=11$ and $2a\sqrt{b}=2\sqrt{30}$
$\Rightarrow {{a}^{2}}+b=11.........\left( i \right)$ and $a=\dfrac{\sqrt{30}}{\sqrt{b}}.........\left( ii \right)$
From equation (ii), we will put the value of a in equation (i). So, we get,
$\begin{align}
& {{\left( \dfrac{\sqrt{30}}{\sqrt{b}} \right)}^{2}}+b=11 \\
& \dfrac{30}{b}+b=11 \\
& \dfrac{30+{{b}^{2}}}{b}=11 \\
& 30+{{b}^{2}}=11b \\
& {{b}^{2}}-11b+30=0 \\
\end{align}$
And, by middle term spitting method, we can write,
$\begin{align}
& {{b}^{2}}-5b-6b+30=0 \\
& b\left( b-5 \right)-6\left( b-5 \right)=0 \\
& \left( b-5 \right)\left( b-6 \right)=0 \\
\end{align}$
$b-5=0$ or $b-6=0$
b = 5 or b = 6
Now, we will put both the values of b one by one in equation (ii). So, we get,
For b = 5, $a=\dfrac{\sqrt{30}}{\sqrt{5}}=\sqrt{6}$ and for b = 6, $a=\dfrac{\sqrt{30}}{\sqrt{6}}=\sqrt{5}$.
Therefore, for b = 5 and $a=\sqrt{6}$, we get $\left( a+\sqrt{b} \right)=\left( \sqrt{6}+\sqrt{5} \right)$ and, for b = 6 and $a=\sqrt{5}$, we get $\left( a+\sqrt{b} \right)=\left( \sqrt{5}+\sqrt{6} \right)$. In both the cases we got \[\left( a+\sqrt{b} \right)=\left( \sqrt{5}+\sqrt{6} \right)\]. Hence, we can say that the square root of $11+2\sqrt{30}$ is \[\left( \sqrt{5}+\sqrt{6} \right)\].
Note: We can also solve this question by hit and trial method, by writing $\sqrt{30}=\sqrt{2}\times \sqrt{15}=\sqrt{6}\times \sqrt{5}=1\times \sqrt{30}$ and then by choosing the correct possible pair, which might give us the correct answer for this small value. But if we get some bigger value and then apply the hit and trial method, it will waste a lot of time. So, it is better to solve this question using the conventional method.
Recently Updated Pages
Write any three differences between metals and nonmetals class 10 social science CBSE

Amit standing on a horizontal plane finds a bird flying class 10 maths CBSE

Two circles of radii 5 cm and 3 cm intersect at two class 10 maths CBSE

Solve the following i John and Jivanti together have class 10 maths CBSE

What is the relation between orthocenter circumcentre class 10 maths CBSE

Two plane mirrors are inclined at 70circ A ray incident class 10 physics CBSE

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Which country is known as "The land of Fire and Ice"?

10 examples of evaporation in daily life with explanations

What is the full form of POSCO class 10 social science CBSE

Make a sketch of the human nerve cell What function class 10 biology CBSE

Choose the feminine form of the given noun Fox AFoxess class 10 english CBSE

