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December 20, 2014

December 20, 2014

Posted by **Test** on Monday, June 17, 2013 at 5:19am.

before we do anything, remember we cannot take the square root of a negative so,

5x-4â‰¥0 and 13-xâ‰¥0

xâ‰¥4/5 and x â‰¤ 13

now back to the question,

square both sides

5x-4 = 49 - 14âˆš(13-x) + 13-x

14âˆš(13-x) = 66 - 6x

7âˆš(13-x) = 33 - 3x

square it again:

49(13-x) = 1089 - 198x + 9x^2

9x^2-149x + 452 = 0

x = (149 Â± âˆš5929)/18

= (149 Â± 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 = âˆš(20-4) = âˆš16 = 4

RS = 7 - âˆš9 = 7-3 = 4

x = 4

- Check on Test : Test -
**Reiny**, Monday, June 17, 2013 at 7:44amThis 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

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