SOLUTION: I need to find x, AB, and BC, where A, B, and C are points on a line...A is the leftmost point, C is the rightmost point, and B is in between A and C (not necessarily the midway)

Algebra ->  Radicals -> SOLUTION: I need to find x, AB, and BC, where A, B, and C are points on a line...A is the leftmost point, C is the rightmost point, and B is in between A and C (not necessarily the midway)       Log On


   

Question 156831: I need to find x, AB, and BC, where A, B, and C are points on a line...A is the leftmost point, C is the rightmost point, and B is in between A and C (not necessarily the midway)
AB = square root (3x +4)
BC = x-6
AC = 6
I was able to write out sqrt(3x+4) + x - 6 = 6, then simplifies to SQRT (3x+4) + x = 12. I don't understand how to eliminate the square root in these radical equations.
Please help.
Thank you.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

sqrt%283x%2B4%29+%2B+x+-6=6 Start with the given equation.


sqrt%283x%2B4%29+%2B+x+=12 Add 6 to both sides.


sqrt%283x%2B4%29++=-x%2B12 Subtract "x" from both sides.


3x%2B4++=%28-x%2B12%29%5E2 Square both sides to eliminate the square root


3x%2B4++=x%5E2-24x%2B144 FOIL


0=x%5E2-24x%2B144-3x-4 Subtract 3x from both sides. Subtract 4 from both sides.


0=x%5E2-27x%2B140 Combine like terms.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=1, b=-27, and c=140


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%28-27%29+%2B-+sqrt%28+%28-27%29%5E2-4%281%29%28140%29+%29%29%2F%282%281%29%29 Plug in a=1, b=-27, and c=140


x+=+%2827+%2B-+sqrt%28+%28-27%29%5E2-4%281%29%28140%29+%29%29%2F%282%281%29%29 Negate -27 to get 27.


x+=+%2827+%2B-+sqrt%28+729-4%281%29%28140%29+%29%29%2F%282%281%29%29 Square -27 to get 729.


x+=+%2827+%2B-+sqrt%28+729-560+%29%29%2F%282%281%29%29 Multiply 4%281%29%28140%29 to get 560


x+=+%2827+%2B-+sqrt%28+169+%29%29%2F%282%281%29%29 Subtract 560 from 729 to get 169


x+=+%2827+%2B-+sqrt%28+169+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%2827+%2B-+13%29%2F%282%29 Take the square root of 169 to get 13.


x+=+%2827+%2B+13%29%2F%282%29 or x+=+%2827+-+13%29%2F%282%29 Break up the expression.


x+=+%2840%29%2F%282%29 or x+=++%2814%29%2F%282%29 Combine like terms.


x+=+20 or x+=+7 Simplify.


So the possible answers are x+=+20 or x+=+7

However, if you plug x+=+20 back into the original equation, you'll find that it doesn't work.


So the only solution is x+=+7



Now plug in x+=+7 into AB, BC, and AC


AB=sqrt%283x%2B4%29=sqrt%283%287%29%2B4%29=sqrt%2821%2B4%29=sqrt%2825%29=5
BC=x-6=%287%29-6=1
AC=6


So the lengths are AB=5, BC=1, and AC=6

Check:

Remember, AB+BC=AC

AB+BC=AC ... Start with the given equation

5+1=6 ... Plug in AB=5, BC=1, and AC=6

6=6 ... Add Since this equation is true, this verifies the answer.