SOLUTION: I'm trying to solve the following equation: The square root of [x+10] plus the square root of [x-6] equals 8. I tried squaring both sides, and that eliminates some radicals, but

Algebra ->  Radicals -> SOLUTION: I'm trying to solve the following equation: The square root of [x+10] plus the square root of [x-6] equals 8. I tried squaring both sides, and that eliminates some radicals, but       Log On


   

Question 118200: I'm trying to solve the following equation:
The square root of [x+10] plus the square root of [x-6] equals 8.
I tried squaring both sides, and that eliminates some radicals, but it also creates some more!
Found 2 solutions by ankor@dixie-net.com, bucky:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The square root of [x+10] plus the square root of [x-6] equals 8.
:
sqrt%28x%2B10%29 + sqrt%28x-6%29 = 8
:
Subtract sqrt%28x-6%29 from both sides:
sqrt%28x%2B10%29 = 8 - sqrt%28x-6%29
:
Square both sides, (FOIL the right side)
x + 10 = 64 - 8sqrt%28x-6%29 - sqrt%28x-6%29 + (x-6)
:
x + 10 = 64 - 6 - 16sqrt%28x-6%29 + x
:
x + 10 = 58 + x - 16sqrt%28x-6%29
:
x - x + 10 - 58 = 16sqrt%28x-6%29
:
-48 = -16x-6%29
:
Square the equation again:
2304 = 256(x-6)
:
2304 = 256x - 1536
:
256x = 2304 + 1536
:
256x = 3840
:
x = 3840/256
:
x = 15
:
:
Check in the original equation
sqrt%2815%2B10%29 + sqrt%2815-6%29 = 8

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
sqrt%28x%2B10%29+%2B+sqrt%28x-6%29+=+8
.
If you have two radicals involved, it often simplifies things to have one radical on one side
of the equal sign and the other radical on the other side. So let's begin by subtracting
sqrt%28x-6%29 from both sides. When you do that the equation becomes:
.
sqrt%28x+%2B+10%29+=+8+-+sqrt%28x-6%29
.
Now square both sides. The left side becomes just x + 10. And on the right side you multiply:
.
%288+-+sqrt%28x-6%29%29%2A%288+-+sqrt%28x+-+6%29%29
.
Without going into the detail of multiplying these two binomials, the product is:
.
64+-+16%2Asqrt%28x-6%29+%2B+%28x-6%29
.
Substituting these new left and right sides into the equation results in:
.
x%2B10+=+64+-+16%2Asqrt%28x-6%29+%2B+x+-6
.
Subtract x from both sides of the equation ... which gets rid of the two x terms and leaves:
.
10+=+64+-+16%2Asqrt%28x-6%29+-6
.
On the right side combine the +64 and the -6 to make the right side become:
.
10+=+-16%2Asqrt%28x+-+6%29+%2B+58
.
Get rid of the 10 on the left side by subtracting 10 from both sides:
.
0+=+-16%2Asqrt%28x-6%29+%2B+48
.
Get rid of the radical on the right side by adding 16%2Asqrt%28x-6%29 to both sides and you
have:
.
16%2Asqrt%28x-6%29+=+48
.
Divide both sides by 16 and you reduce the equation to:
.
sqrt%28x-6%29+=+3
.
Square both sides to get:
.
x+-+6+=+9
.
Finally, solve for x by adding 6 to both sides to get rid of the -6 on the left side. When
you do that, you have the answer:
.
x+=+15
.
Check this answer by going back to the original equation:
.
sqrt%28x%2B10%29+%2B+sqrt%28x-6%29+=+8
.
Substitute 15 for x and it becomes:
.
sqrt%2815%2B10%29+%2B+sqrt%2815-6%29+=+8
.
and this reduces to:
.
sqrt%2825%29+%2B+sqrt%289%29+=+8
.
Taking the two square roots gives you:
.
5+%2B+3+=+8
.
That works ... so the answer of y = 15 looks good.
.
Hope this helps you to understand the problem and how to solve it. Splitting the radicals ...
one on one side and the other on the other side seemed to help a bit.
.