How do you solve #3 - x =sqrt(x^2 + 15)#?

1 Answer
Aug 11, 2015

Answer:

#x = -1#

Explanation:

Start by taking a look at the expression that's under the square root.

#x^2 + 15>0, (AA) x in RR#

In other words, as far as the radical term is concerned, #x# can take any value in #RR#. On the other hand, the possible values of #x# are restricted by the expression on the left side of the equation.

Since the square root of a positive number is always a positive number, you have

  • #3-x>=0 implies x<=3#

Now, square both sides of the equation to get rid of the radical term

#(3-x)^2 = (sqrt(x^2 + 15))^2#

#9 - 6x + color(red)(cancel(color(black)(x^2))) = color(red)(cancel(color(black)(x^2))) + 15#

This is equivalent to

#-6x = 6 implies x= 6/(-6) = color(green)(-1)#

Since #x=-1# satisfies the condition #x<=3#, this will be the solution to the original equation.

Do a quick check to make sure that everything came out right

#3 - (-1) = sqrt( (-1)^2 + 15)#

#3 + 1 = sqrt(16)#

#4 = 4 color(green)(sqrt())#