SOLUTION: I need to resolve x, AB, BC and AC, where A, B and C are points on a line. A is leftmost, C is rightmost, and B is somewhere in between. AB = 2x+1 BC = x-2 AC = sqrt (x^2 + 2

Algebra ->  Algebra  -> Radicals -> SOLUTION: I need to resolve x, AB, BC and AC, where A, B and C are points on a line. A is leftmost, C is rightmost, and B is somewhere in between. AB = 2x+1 BC = x-2 AC = sqrt (x^2 + 2      Log On

Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   

Question 156834: I need to resolve x, AB, BC and AC, where A, B and C are points on a line. A is leftmost, C is rightmost, and B is somewhere in between.
AB = 2x+1
BC = x-2
AC = sqrt (x^2 + 25x + 5)
Was able to write, 2x+1+x-2 = sqrt (x^2+25x+5), which reduces to 3x-1 = sqrt (x^2 + 25x + 5).
I don't know how to eliminate the square root, especially when there is an exponential along with it.
Please help.
Thank you.
Answer by jim_thompson5910(21667) About Me  (Show Source):
You can put this solution on YOUR website!
I'll start where you left off

3x-1+=+sqrt+%28x%5E2+%2B+25x+%2B+5%29 Start with the given equation


%283x-1%29%5E2+=+x%5E2+%2B+25x+%2B+5 Square both sides. This will eliminate the square root.


9x%5E2-6x%2B1+=+x%5E2+%2B+25x+%2B+5 FOIL the left side


9x%5E2-6x%2B1-x%5E2-25x-5=0 Subtract x%5E2 from both sides. Subtract 25x from both sides. Subtract 5 from both sides.


8x%5E2-31x-4=0 Combine like terms.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=8, b=-31, and c=-4


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-%28-31%29+%2B-+sqrt%28+%28-31%29%5E2-4%288%29%28-4%29+%29%29%2F%282%288%29%29 Plug in a=8, b=-31, and c=-4


x+=+%2831+%2B-+sqrt%28+%28-31%29%5E2-4%288%29%28-4%29+%29%29%2F%282%288%29%29 Negate -31 to get 31.


x+=+%2831+%2B-+sqrt%28+961-4%288%29%28-4%29+%29%29%2F%282%288%29%29 Square -31 to get 961.


x+=+%2831+%2B-+sqrt%28+961--128+%29%29%2F%282%288%29%29 Multiply 4%288%29%28-4%29 to get -128


x+=+%2831+%2B-+sqrt%28+961%2B128+%29%29%2F%282%288%29%29 Rewrite sqrt%28961--128%29 as sqrt%28961%2B128%29


x+=+%2831+%2B-+sqrt%28+1089+%29%29%2F%282%288%29%29 Add 961 to 128 to get 1089


x+=+%2831+%2B-+sqrt%28+1089+%29%29%2F%2816%29 Multiply 2 and 8 to get 16.


x+=+%2831+%2B-+33%29%2F%2816%29 Take the square root of 1089 to get 33.


x+=+%2831+%2B+33%29%2F%2816%29 or x+=+%2831+-+33%29%2F%2816%29 Break up the expression.


x+=+%2864%29%2F%2816%29 or x+=++%28-2%29%2F%2816%29 Combine like terms.


x+=+4 or x+=+-1%2F8 Simplify.


So the possible answers are x+=+4 or x+=+-1%2F8

However, if you plug in x+=+-1%2F8 back into the original equation, you'll find that the equation won't be true. Also, since distance is ALWAYS positive, this means that x+=+-1%2F8 will not work (since if you plug it into any expression the result is negative)


So the only answer is x+=+4


Now plug in x=4 into AB,BC, and AC:

AB = 2x%2B1=2%284%29%2B1=8%2B1=9
BC = x-2=%284%29-2=2
AC = sqrt+%28x%5E2+%2B+25x+%2B+5%29+=sqrt+%28%284%29%5E2+%2B+25%284%29+%2B+5%29=sqrt+%2816+%2B+100+%2B+5%29=sqrt+%28121%29=11


So the three lengths are AB=9, BC=2, and AC=11

Check:

Remember, the segment addition postulate is AB+BC=AC (ie the lengths of the pieces of AC should add to the length of AC)

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

9+2=11 ...Plug in AB=9, BC=2, and AC=11

11=11 ... Add. Since this equation is true, this verifies the answer.