What is \[\sqrt {126} + \sqrt {56} \]in the standard form?
Answer
561k+ views
Hint: To write the given expression in standard form, we need to simplify both the terms in the expression first. For that, first we need to prime factorise the numbers and then write the numbers which can be written as squares in the form of squares of that number and then find their square root and write their square root in simplified form. We know,
\[x \times x = {x^2}\]
\[\sqrt {{x^2}} = |x|\]
So, in this question, when we square root the prime factors, we will use the above formula and then multiply it with the remaining terms. Also, we know
\[\sqrt {xy} = \sqrt x \sqrt y \]
\[x\sqrt y + z\sqrt y = (x + z)\sqrt y \]
\[x\sqrt y + w\sqrt z \] cannot be added.
And, we will use this to add the simplified form of the two numbers.
Complete step-by-step solution:
We need to find \[\sqrt {126} + \sqrt {56} \]in standard form.
For that, we will simplify \[\sqrt {126} \]and \[\sqrt {56} \]separately and then add them.
Prime Factorisation \[126\], we get,
\[126 = 2 \times 3 \times 3 \times 7\]
\[126 = 2 \times {3^2} \times 7\]
Now, square rooting both the sides,
\[\sqrt {126} = \sqrt {2 \times {3^2} \times 7} \]
\[\sqrt {126} = \sqrt {2 \times 7} \times \sqrt {{3^2}} \] (Separating the square terms)
\[\sqrt {126} = \sqrt {2 \times 7} \times |3|\]
\[\sqrt {126} = \sqrt {14} \times 3\]
\[\sqrt {126} = 3\sqrt {14} \] ----(1)
Now, simplifying \[\sqrt {56} \]
Prime Factorisation \[56\], we get,
\[56 = 2 \times 2 \times 2 \times 7\]
\[56 = {2^2} \times 2 \times 7\] (Writing \[2 \times 2\] as a \[{2^2}\])
Square rooting both sides,
\[\sqrt {56} = \sqrt {{2^2} \times 2 \times 7} \]
\[\sqrt {56} = \sqrt {{2^2}} \times \sqrt {2 \times 7} \] (Separating the square term)
\[\sqrt {56} = |2| \times \sqrt {2 \times 7} \]
\[\sqrt {56} = 2 \times \sqrt {14} \]
\[\sqrt {56} = 2\sqrt {14} \] -----(2)
Now, from (1) and (2), we get,
\[\sqrt {126} + \sqrt {56} = 3\sqrt {14} + 2\sqrt {14} \]
\[\sqrt {126} + \sqrt {56} = 5\sqrt {14} \]
Hence, \[\sqrt {126} + \sqrt {56} = 5\sqrt {14} \]in standard form.
Note: While solving, we need to keep in mind that the terms which cannot be written as squares will remain in under root and the ones which can be written as squares will be simplified. And, we need to make sure that terms in square root can be added and what terms cannot be. In these types of questions, when we take square root, we usually don’t take negative signs but only positive signs.
\[x \times x = {x^2}\]
\[\sqrt {{x^2}} = |x|\]
So, in this question, when we square root the prime factors, we will use the above formula and then multiply it with the remaining terms. Also, we know
\[\sqrt {xy} = \sqrt x \sqrt y \]
\[x\sqrt y + z\sqrt y = (x + z)\sqrt y \]
\[x\sqrt y + w\sqrt z \] cannot be added.
And, we will use this to add the simplified form of the two numbers.
Complete step-by-step solution:
We need to find \[\sqrt {126} + \sqrt {56} \]in standard form.
For that, we will simplify \[\sqrt {126} \]and \[\sqrt {56} \]separately and then add them.
Prime Factorisation \[126\], we get,
\[126 = 2 \times 3 \times 3 \times 7\]
\[126 = 2 \times {3^2} \times 7\]
Now, square rooting both the sides,
\[\sqrt {126} = \sqrt {2 \times {3^2} \times 7} \]
\[\sqrt {126} = \sqrt {2 \times 7} \times \sqrt {{3^2}} \] (Separating the square terms)
\[\sqrt {126} = \sqrt {2 \times 7} \times |3|\]
\[\sqrt {126} = \sqrt {14} \times 3\]
\[\sqrt {126} = 3\sqrt {14} \] ----(1)
Now, simplifying \[\sqrt {56} \]
Prime Factorisation \[56\], we get,
\[56 = 2 \times 2 \times 2 \times 7\]
\[56 = {2^2} \times 2 \times 7\] (Writing \[2 \times 2\] as a \[{2^2}\])
Square rooting both sides,
\[\sqrt {56} = \sqrt {{2^2} \times 2 \times 7} \]
\[\sqrt {56} = \sqrt {{2^2}} \times \sqrt {2 \times 7} \] (Separating the square term)
\[\sqrt {56} = |2| \times \sqrt {2 \times 7} \]
\[\sqrt {56} = 2 \times \sqrt {14} \]
\[\sqrt {56} = 2\sqrt {14} \] -----(2)
Now, from (1) and (2), we get,
\[\sqrt {126} + \sqrt {56} = 3\sqrt {14} + 2\sqrt {14} \]
\[\sqrt {126} + \sqrt {56} = 5\sqrt {14} \]
Hence, \[\sqrt {126} + \sqrt {56} = 5\sqrt {14} \]in standard form.
Note: While solving, we need to keep in mind that the terms which cannot be written as squares will remain in under root and the ones which can be written as squares will be simplified. And, we need to make sure that terms in square root can be added and what terms cannot be. In these types of questions, when we take square root, we usually don’t take negative signs but only positive signs.
Recently Updated Pages
What are the methods of reducing friction. Explain

Find the largest number of six digits which is a p-class-8-maths-CBSE

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

What does the color green in the national flag of India class 8 social science CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Full form of STD, ISD and PCO

One cusec is equal to how many liters class 8 maths CBSE

Which Indian state shares the longest international class 8 social science CBSE


