Solve - radical x+2 = x-4?

Answer 1

sqrt(x+2) = x-4
x+2 = x^2 -8x+16
x^2 - 9x +14 = 0
(x-2)(x-7) = 0
so x = 2 or 7

Answer 2

I suppose you mean
-sqrt(x+2) =x-4
1) x+2>=0 so x>=-2
2) x-4<=0 as the f
left side is negative or 0 x<=4
With this conditions sqare both sides
x+2 =x^2-8x+16
x^2-9x+14=0
x=((9+-sqrt(81-56))/2The positive sign gives a root x= 7 outside the range
so x= 2 inside the range

Answer 3

Square both sides, and check the solutions:

x + 2 = (x - 4)^2

x + 2 = x^2 - 8x + 16 // subtract x + 2

x^2 - 9x + 14 = 0

x^2 - 2x * 4.5 + 4.5^2 - 4.5^2 + 14 = 0

(x - 4.5)^2 - 20.25 + 14 = 0

(x - 4.5)^2 - 6.25 = 0

(x - 4.5)^2 - 2.5^2 = 0

(x - 4.5 - 2.5)(x - 4.5 + 2.5)=0

(x - 7)(x - 2) = 0

Let's check x = 7

radical (7 + 2) = radical 9 = 3 = 7 -4

Let's check x = 2

radical (2 + 2) = radical 4 = 2, but 2 - 4 = -2

So, we only have x=7

Answer 4

sqrt(x + 2) = x - 4

We've gotta be careful with this one, because whenever we square something (which we need to do to get rid of the sqrt) we may introduce an extraneous solution (false "solution")

so first we square each side to get

x + 2 = (x - 4)^2
x + 2 = x^2 - 8x + 16
subtracting x and 2 from the left side we get
0 = x^2 - 9c + 14
We need 2 factors that multiply to get 14 and add to get -9... those factors are -2 and -7, so we get
0 = (x - 2)(x - 7)
so, x - 2 = 0 or x - 7 = 0
solving these we get x = 2, 7
We need to check our answers to see if we got an extraneous solution...
when we plug 7 in the equation we get
sqrt(7 + 2) = 7 - 4
sqrt(9) = 3
3 = 3
so we're good...
when we plug 2 into the equation we get
sqrt(2+2) = 2 - 4
sqrt(4) = -2
2 = -2
so we got an extraneous solution...
so the correct answer is x = 7.

Answer 5

Radical x+2=x-4

square both sides

x+2=(x-4)^2
x+2=x^2-8x+16
x-x+2-2=x^2-8x-x+16=2
x^2-9x-14=0
(x-2)(x-7)=0
x=2 x=7

Answer 6

x+2=x^2+16-8x

x^2+16-8x-x-2=0

x^2-9x+14=0

x1=2 does not satisfy

x2=7 is the answer