How do you solve #-4 sqrt(x+9) = 20# and find any extraneous solutions?

1 Answer
Aug 3, 2016

Answer:

#x in O/#

Explanation:

Your radical equation has no solutions for real numbers. Here's why.

Isolate the square root on one side of the equation by dividing both sides by #-4#

#(color(red)(cancel(color(black)(-4))) * sqrt(x+9))/color(red)(cancel(color(black)(-4))) = 20/(-4)#

#sqrt(x+9) = -5#

By definition, the square root of a positive real number must always produce a positive value. This implies that you must have

  • #x + 9 >= 0 -># you must take the square root of a positive number when working in #RR#

  • #color(red)(cancel(color(black)(-5 >=0))) -># the square root of a positive number must be a positive number

In your case ,#-5# is clearly not positive, which implies that your equation has no solutions when working in #RR#. Simply put, you can't find a value of #x in RR# that can satisfy both conditions.

You can have #x+9 >=0# for #x in [-9, +oo)#, but you will never have #-5 >=0#, regardless of the value of #x in [-9, + oo)#.

You can write this as

#x in O/ -># no solutions in #RR#