How do you solve #sqrt(x+15) - sqrt(2x+7)=1#?

1 Answer
Aug 11, 2015

Answer:

#x = 1#

Explanation:

Even before doing any calculations, you know that any possible solution to this equation must satisfy two conditions

  • #x+15>=0 implies x>= - 15#
  • #2x+7>=0 implies x >= -7/2#
  • #sqrt(x+15)>sqrt(2x+7)#

The third condition implies that #x+15>2x+7 implies x<8#.

Combining these conditions will result in #x in [-7/2, 8)#.

These condtions are required because you can't take the square root of a negative number if you're working with real numbers.

In other words, you can only take the square root of positive real numbers; likewise, taking the square root of a positive number will always result in a positive number.

Start by squaring both sides of the equation; this will reduce the number of radical terms from two to one

#(sqrt(x+15) - sqrt(2x+7))^2 = 1^2#

#(sqrt(x+15))^2 - 2sqrt((x+15)(2x+7)) + (sqrt(2x+7))^2 = 1#

#x + 15 - 2sqrt((x+15)(2x+7)) + 2x + 7 = 1#

Isolate the remaining radical term on one side of the equation

#-2sqrt((x+15)(2x+7)) = 1 - 3x - 22#

#2sqrt((x+15)(2x+7)) = 3x + 21#

Once again, square both sides of the equation to get rid of the radical lterm

#(2sqrt((x+15)(2x+7)))^2 = (3x+21)^2#

#4(x+15)(2x+7) = 9x^2 + 126x + 441#

#8x^2 + 148x + 420 = 9x^2 +126x + 441#

Rearrange this equation to classic quadratic form

#x^2 - 22x + 21 = 0#

You can use the quadratic equation to find the two roots of this quadratic equation

#x_(1,2) = (-(-22) +- sqrt((-22)^2 - 4 * 1 * 21))/(2 * 1)#

#x_(1,2) = (22 +- sqrt(400))/2 = (22 +- 20)/2 = {(x_1 = (22 + 20)/2 = 21), (x_2 = (22-20)/2 = 1) :}#

Notice that #x_1# does not satisfy the condition #x in [-7/2, 8)#, which means that the only valid solution to the original equation is #x_2=1#.

More specifically,

#sqrt(21 + 15) - sqrt(2 * 21 + 7) = 1#

#6 - 7 color(red)(!=) 1 -> x_1# is an extraneous solution

#sqrt(1 + 15) - sqrt(2 * 1 + 7) = 1#

#4 - 3 = 1 color(green)(sqrt())#