How do you solve $sqrt(x + 6) + sqrt(2-x) = 4$ ?

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Answer 1 · 2015-10-07T00:28:40

I found $x=-2$

Let us try squaring both sides:

$[sqrt(x+6)+sqrt(2-x)]^2=4^2$

$cancel(x)+6+2sqrt(x+6)sqrt(2-x)+2cancel(-x)=16$

$8+2sqrt((x+6)(2-x))=16$ taking $8$ to the right:

$2sqrt((x+6)(2-x))=8$ taking $2$ to the right (dividing):

$sqrt((x+6)(2-x))=4$

let us square again:

$(x+6)(2-x)=16$

$2x-x^2+12-6x-16=0$

$-x^2-4x-4=0$ using the Quadratic Formula we get:

$x_(1,2)=(4+-sqrt(16-4(-4*-1)))/-2=(4+-sqrt(0))/-2=-2$

Let us try this solution into our original equation as $x$ :
$sqrt(-2+6)+sqrt(2+2)=2+2=4$ Yes.

How do you solve sqrt(x + 6) + sqrt(2-x) = 4sqrt(x + 6) + sqrt(2-x) = 4?