How do you simplify #5sqrt7-4sqrt7#?

1 Answer
Ryan · MathFact-orials.blogspot.com
Jun 21, 2018

Hello you can simplify this by recognizing that '4' and '5' are coefficients of the radical #sqrt7#

Explanation:

In this it asks #5sqrt7# - #4sqrt7#
Do it by simply subtracting #5-4 = 1#
now it is #sqrt7# or 'one of #sqrt7#' since it is like adding and subtracting like terms but similar with radicals.

To make this more clear, we can view the coefficients in front of the radical as entities on their own or write them this way as they are notated by a coefficient;

#sqrt7 * sqrt7 * sqrt7 * sqrt7 * sqrt7 - sqrt7 * sqrt7 * sqrt7 * sqrt7#

Simply by subtracting 4 from 5, we can get an answer on how many radicals are left in the remainder as notated by #5sqrt7# - #4sqrt7# .

Thus, #5sqrt7# - #4sqrt7=sqrt7#

For general examples;
#asqrtc - bsqrtc = (a-b)sqrtc#

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