Question 156834
I'll start where you left off


{{{3x-1 = sqrt (x^2 + 25x + 5)}}} Start with the given equation



{{{(3x-1)^2 = x^2 + 25x + 5}}} Square both sides. This will eliminate the square root.



{{{9x^2-6x+1 = x^2 + 25x + 5}}} FOIL the left side



{{{9x^2-6x+1-x^2-25x-5=0}}} Subtract {{{x^2}}} from both sides.  Subtract {{{25x}}} from both sides.  Subtract {{{5}}} from both sides.



{{{8x^2-31x-4=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=8}}}, {{{b=-31}}}, and {{{c=-4}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-31) +- sqrt( (-31)^2-4(8)(-4) ))/(2(8))}}} Plug in  {{{a=8}}}, {{{b=-31}}}, and {{{c=-4}}}



{{{x = (31 +- sqrt( (-31)^2-4(8)(-4) ))/(2(8))}}} Negate {{{-31}}} to get {{{31}}}. 



{{{x = (31 +- sqrt( 961-4(8)(-4) ))/(2(8))}}} Square {{{-31}}} to get {{{961}}}. 



{{{x = (31 +- sqrt( 961--128 ))/(2(8))}}} Multiply {{{4(8)(-4)}}} to get {{{-128}}}



{{{x = (31 +- sqrt( 961+128 ))/(2(8))}}} Rewrite {{{sqrt(961--128)}}} as {{{sqrt(961+128)}}}



{{{x = (31 +- sqrt( 1089 ))/(2(8))}}} Add {{{961}}} to {{{128}}} to get {{{1089}}}



{{{x = (31 +- sqrt( 1089 ))/(16)}}} Multiply {{{2}}} and {{{8}}} to get {{{16}}}. 



{{{x = (31 +- 33)/(16)}}} Take the square root of {{{1089}}} to get {{{33}}}. 



{{{x = (31 + 33)/(16)}}} or {{{x = (31 - 33)/(16)}}} Break up the expression. 



{{{x = (64)/(16)}}} or {{{x =  (-2)/(16)}}} Combine like terms. 



{{{x = 4}}} or {{{x = -1/8}}} Simplify. 



So the possible answers are {{{x = 4}}} or {{{x = -1/8}}} 

  

However, if you plug in {{{x = -1/8}}} back into the original equation, you'll find that the equation won't be true. Also, since distance is <b>ALWAYS</b> positive, this means that {{{x = -1/8}}} 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+1=2(4)+1=8+1=9}}}
BC = {{{x-2=(4)-2=2}}}
AC = {{{sqrt (x^2 + 25x + 5) =sqrt ((4)^2 + 25(4) + 5)=sqrt (16 + 100 + 5)=sqrt (121)=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.