How do you solve #\sqrt { x + 6} = 6- 2\sqrt { 5- x }#?

1 Answer
Aug 20, 2017

Answer:

#x ~~ 2.66#

Explanation:

For starters, you know that you must have

#x + 6 >= 0 implies x >= - 6#

and

#5 - x >= 0 implies x <=5 #

That is the case because the expressions under the two square roots must be positive when working with real numbers.

So, right from start, you know that the solution interval cannot include any value of #x# that is not part of

#x in [-6, 5]" "color(blue)((1))#

Moreover, you need to have

#6 - 2sqrt(5-x) >= 0#

#6 >= 3 sqrt(5-x)#

This simplifies to

#2 >= sqrt(5-x)" "color(blue)((2))#

Now, rearrange the equation as

#sqrt(x+6) + 2sqrt(5-x) = 6#

Square both sides of the equation to get

#(sqrt(x+6) + 2 sqrt(5-x))^2 = 6^2#

#(sqrt(x+6))^2 + 2 * sqrt(x+6) * 2 * sqrt(5-x) + (2sqrt(5-x))^2 = 36#

This is equivalent to

#x + 6 + 4 sqrt((x+5)(5-x)) + 4(5-x) = 36#

Rearrange the equation to isolate the new square root term on one side of the equation

#4sqrt((x+6)(5-x)) = 36 - x - 6 - 20 + 4x#

#sqrt((x+6)(x-5)) = (10 + 3x)/4#

Next, square both sides of the equation again to get

#(sqrt((x+6)(x-5)))^2 = ((10 +3x)/4)^2#

#(x+6)(5-x) = (100 + 60x + 9x^2)/16#

This will be equivalent to

#-x^2 - x + 30 = (100 + 60x + 9x^2)/16#

#-16x^2 - 16x + 480 = 9x^2 + 60x + 100#

Rearrange to quadratic equation form

#25x^2 + 76x -380 = 0#

Use the quadratic formula to find the two roots of this quadratic equation

#x_ (1,2) = (-76 +- sqrt(76^2 - 4 * 25 * (-380)))/(2 * 25)#

#x_(1,2) ~~ (-76 +- 209)/50 implies {(x_1 ~~ (-76 - 209)/50 ~~ -5.7), (x_2 ~~ (-76 + 209)/50 ~~ 2.66) :}#

Now, notice that both solutions satisfy #color(blue)((1))#, but that only one satisfies #color(blue)((2))#, since

#2 color(red)(cancel(color(black)(>=))) sqrt(5 - (-5.7))#

#2 color(red)(cancel(color(black)(>=))) sqrt(10.7)#

This means that #x ~~ -5.7# is an extraneous solution, i.e. it does not satisfy the original equation.

Therefore, you can say that #x ~~ 2.66# is the only solution to the original equation.