How do you solve \sqrt { x + 6} = 6- 2\sqrt { 5- x }√x+6=6−2√5−x?
1 Answer
Explanation:
For starters, you know that you must have
x + 6 >= 0 implies x >= - 6x+6≥0⇒x≥−6
and
5 - x >= 0 implies x <=5 5−x≥0⇒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))x∈[−6,5] (1)
Moreover, you need to have
6 - 2sqrt(5-x) >= 06−2√5−x≥0
6 >= 3 sqrt(5-x)6≥3√5−x
This simplifies to
2 >= sqrt(5-x)" "color(blue)((2))2≥√5−x (2)
Now, rearrange the equation as
sqrt(x+6) + 2sqrt(5-x) = 6√x+6+2√5−x=6
Square both sides of the equation to get
(sqrt(x+6) + 2 sqrt(5-x))^2 = 6^2(√x+6+2√5−x)2=62
(sqrt(x+6))^2 + 2 * sqrt(x+6) * 2 * sqrt(5-x) + (2sqrt(5-x))^2 = 36(√x+6)2+2⋅√x+6⋅2⋅√5−x+(2√5−x)2=36
This is equivalent to
x + 6 + 4 sqrt((x+5)(5-x)) + 4(5-x) = 36x+6+4√(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 + 4x4√(x+6)(5−x)=36−x−6−20+4x
sqrt((x+6)(x-5)) = (10 + 3x)/4√(x+6)(x−5)=10+3x4
Next, square both sides of the equation again to get
(sqrt((x+6)(x-5)))^2 = ((10 +3x)/4)^2(√(x+6)(x−5))2=(10+3x4)2
(x+6)(5-x) = (100 + 60x + 9x^2)/16(x+6)(5−x)=100+60x+9x216
This will be equivalent to
-x^2 - x + 30 = (100 + 60x + 9x^2)/16−x2−x+30=100+60x+9x216
-16x^2 - 16x + 480 = 9x^2 + 60x + 100−16x2−16x+480=9x2+60x+100
Rearrange to quadratic equation form
25x^2 + 76x -380 = 025x2+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)x1,2=−76±√762−4⋅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
Therefore, you can say that
