Question

√(5x-4) = 7 - √(13-x)
before we do anything, remember we cannot take the square root of a negative so,
5x-4≥0 and 13-x≥0
x≥4/5 and x ≤ 13


now back to the question,
square both sides


5x-4 = 49 - 14√(13-x) + 13-x
14√(13-x) = 66 - 6x
7√(13-x) = 33 - 3x


square it again:
49(13-x) = 1089 - 198x + 9x^2
9x^2-149x + 452 = 0
x = (149 ± √5929)/18
= (149 ± 77)/18
x = 95 or 4
We can rule out the x=95 since it falls outside our domain established above
So let's test x=4, (answers must be verified after a squaring process)


if x = 4
LS = √(20-4) = √16 = 4
RS = 7 - √9 = 7-3 = 4


x = 4

Answer 1

This looks like the reply I tried to send yesterday and wouldn't work.
My square roots signs, which didn't come our correctly are done on my Mac by using " Option V "
This has worked for me for years.
Here is the original reply I tried posting


√(5x-4) = 7 - √(13-x)
before we do anything, remember we cannot take the square root of a negative so,
5x-4≥0 and 13-x≥0
x≥4/5 and x ≤ 13

now back to the question,
square both sides

5x-4 = 49 - 14√(13-x) + 13-x
14√(13-x) = 66 - 6x
7√(13-x) = 33 - 3x

square it again:
49(13-x) = 1089 - 198x + 9x^2
9x^2-149x + 452 = 0
x = (149 ± √5929)/18
= (149 ± 77)/18
x = 95 or 4
We can rule out the x=95 since it falls outside our domain established above
So let's test x=4, (answers must be verified after a squaring process)

if x = 4
LS = √(20-4) = √16 = 4
RS = 7 - √9 = 7-3 = 4

x = 4