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

##### 1 Answer

#### Answer:

#### 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 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

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

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

This means that *extraneous solution*, i.e. it does not satisfy the original equation.

Therefore, you can say that