Question

How do you add \[\sqrt 3 + \sqrt {12} - 4\sqrt {75} \] ?

Answer 1

Hint: Here the very first term is our hint to lead the solution of the problem. We have to write the remaining two terms in the form such that the root should have 3 such that 12 is multiple of 4 with product 3 and 75 is a multiple of 25 with product 3 so that the terms can be operated with the given mathematical operations. So let’s start!

Complete step-by-step answer:
Given that,
 \[\sqrt 3 + \sqrt {12} - 4\sqrt {75} \]
We have to write the roots with 3 in the root so go ahead,
 \[ = \sqrt 3 + \sqrt {4 \times 3} - 4\sqrt {25 \times 3} \]
Now we will take the perfect squares outside as their root form, as \[\sqrt 4 = 2\& \sqrt {25} = 5\]
 \[ = \sqrt 3 + 2\sqrt 3 - 4 \times 5\sqrt 3 \]
Taking the product of numbers outside the root,
 \[ = \sqrt 3 + 2\sqrt 3 - 20\sqrt 3 \]
Taking the root common
 \[ = \left( {1 + 2 - 20} \right)\sqrt 3 \]
Now adding the terms we get,
 \[ = - 17\sqrt 3 \]
This is the correct answer.
So, the correct answer is “ \[ = - 17\sqrt 3 \] ”.

Note: Here note that there is no method like we can directly add or subtract the numbers in the root in the situation mentioned above.
 \[\sqrt 3 + \sqrt {12} - 4\sqrt {75} \]
This doesn't allow us to add or subtract any numbers. We need to make the numbers in the root the same. So also no need to take the root common once it is done the same. We can directly add or subtract the number outside the root just keeping the root the same.
 \[ = \sqrt 3 + 2\sqrt 3 - 20\sqrt 3 \]
 \[ = - 17\sqrt 3 \]
This is the same answer.