SOLUTION: square root (x+6)+square root (X-2) = 4 (x+6)2+(x-2)2 X2+6x+6x+36+x2-2x-2x+4=42 2x2+8x+40=16 2x2+8x+24=0 (2x+…)(x+…) I cannot find 2 numbers that equal 24 but also 8 in the

Algebra ->  Rational-functions -> SOLUTION: square root (x+6)+square root (X-2) = 4 (x+6)2+(x-2)2 X2+6x+6x+36+x2-2x-2x+4=42 2x2+8x+40=16 2x2+8x+24=0 (2x+…)(x+…) I cannot find 2 numbers that equal 24 but also 8 in the      Log On


   

Question 150308: square root (x+6)+square root (X-2) = 4
(x+6)2+(x-2)2
X2+6x+6x+36+x2-2x-2x+4=42
2x2+8x+40=16
2x2+8x+24=0
(2x+…)(x+…)
I cannot find 2 numbers that equal 24 but also 8 in the middle... did I approach this exercise wrong? The task was just to solve it...
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
square root (x+6)+square root (X-2) = 4
(x+6)2+(x-2)2
X2+6x+6x+36+x2-2x-2x+4=42
2x2+8x+40=16
2x2+8x+24=0
(2x+…)(x+…)
I cannot find 2 numbers that equal 24 but also 8 in the middle... did I approach this exercise wrong? The task was just to solve it...
------------
sqrt%28x%2B6%29%2Bsqrt%28X-2%29+=+4
Square both sides
%28x%2B6%29+%2B+2%2Asqrt%28x%2B6%29%2Asqrt%28x-2%29+%2B%28x-2%29+=+16
Collect terms
2x+%2B+4+%2B+2%2Asqrt%28x%2B6%29%2Asqrt%28x-2%29+=+16
Isolate radicals
2%2Asqrt%28x%2B6%29%2Asqrt%28x-2%29+=+12+-+2x
Divide by 2
sqrt%28x%2B6%29%2Asqrt%28x-2%29+=+6+-+x
Square both sides again
%28x%2B6%29%2A%28x-2%29+=+x%5E2+-+12x+%2B+36
expand
x%5E2+%2B+4x+-12+=+x%5E2+-12x+%2B+36
Collect terms
16x = 48
x = 3
---- Sub in 3 for x to check
sqrt%28x%2B6%29%2Bsqrt%28X-2%29+=+4
sqrt%283%2B6%29%2Bsqrt%283-2%29+=+4
3 + 1 = 4
Looks good.
You were squaring the binomials inside the radical, shoulda squared just the radical in the first step.