Question

Find the square root of $16-2\sqrt{20}-2\sqrt{28}+2\sqrt{35}$.

Answer 1

Hint: In this question, we are given a number and we have to find its square root. For this, we will simplify the given number into factors. We will find factors in such a way that they form a square of some numbers. So that, when we need to find the square root, the square will get canceled and a number left behind which is our required root. We will use $\sqrt{{{x}^{2}}}=\pm x$. We will use formula as ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz={{\left( x+y+z \right)}^{2}}$.

Complete step-by-step solution:
Here we are given numbers as $16-2\sqrt{20}-2\sqrt{28}+2\sqrt{35}$.
Let us now separate the terms and factorize them. 16 can be written as ${{\left( -2 \right)}^{2}}+5+7$.
$\sqrt{20}$ can be written as $\sqrt{2\times 2\times 5}=2\sqrt{5}$. Hence $-2\sqrt{20}$ can be written as $2\times \left( -2 \right)\times \sqrt{5}$.
$\sqrt{28}$ can be written as $\sqrt{2\times 2\times 7}=2\sqrt{7}$. Hence $-2\sqrt{28}$ can be written as $2\times \left( -2 \right)\times \sqrt{7}$.
$\sqrt{35}$ can be written as $\sqrt{7\times 5}=\sqrt{5}\times \sqrt{7}$. Hence $2\sqrt{35}$ can be written as $2\times \sqrt{5}\times \sqrt{7}$. Combining all terms we get:
\[\Rightarrow {{\left( -2 \right)}^{2}}+5+7+2\times \left( -2 \right)\times \sqrt{5}+2\times \left( -2 \right)\times \sqrt{7}+2\times \sqrt{5}\times \sqrt{7}\]
Here we can write 5 as ${{\left( \sqrt{5} \right)}^{2}}$ and 7 as ${{\left( \sqrt{7} \right)}^{2}}$ we get:
\[\Rightarrow {{\left( -2 \right)}^{2}}+{{\left( \sqrt{5} \right)}^{2}}+{{\left( \sqrt{7} \right)}^{2}}+2\times \left( -2 \right)\times \left( \sqrt{5} \right)+2\times \left( -2 \right)\times \left( \sqrt{7} \right)+2\left( \sqrt{5} \right)\left( \sqrt{7} \right)\cdots \cdots \cdots \left( 1 \right)\]
Now, we know that ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz={{\left( x+y+z \right)}^{2}}$. As we can see equation (1) can be compared with ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz$ so we can write it in the form as ${{\left( x+y+z \right)}^{2}}$ where $x=-2,y=\sqrt{5}\text{ and }z=\sqrt{7}$.
So equation (1) becomes equal to \[\Rightarrow {{\left( -2+\sqrt{5}+\sqrt{7} \right)}^{2}}\].
Now we need to find the square root of the number formed above. As we know, $\sqrt{{{x}^{2}}}=\pm x$ so square root of number becomes
\[\begin{align}
  & \Rightarrow \sqrt{{{\left( -2+\sqrt{5}+\sqrt{7} \right)}^{2}}} \\
 & \Rightarrow \pm \left( -2+\sqrt{5}+\sqrt{7} \right) \\
\end{align}\]
Hence, our required answer is $\pm \left( -2+\sqrt{5}+\sqrt{7} \right)$.

Note: While factoring numbers into ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz$ form, take care of negative signs. Make sure that, both positive and negative are considered after finding the square root of the squared terms. Students can make mistakes in positive and negative signs in these sums so take care of that.