Question 202564
{{{2*sqrt(x-1)-sqrt(3x-5)=sqrt(x-9)}}} Start with the given equation.



{{{(2*sqrt(x-1)-sqrt(3x-5))^2=(sqrt(x-9))^2}}} Square both sides



{{{(2*sqrt(x-1)-sqrt(3x-5))^2=x-9}}} Square the square root.



{{{4(x-1)-4*sqrt(x-1)*sqrt(3x-5)+3x-5=x-9}}} FOIL



{{{4x-4-4*sqrt(x-1)*sqrt(3x-5)+3x-5=x-9}}} Distribute



{{{7x-9-4*sqrt(x-1)*sqrt(3x-5)=x-9}}} Combine like terms.



{{{-4*sqrt(x-1)*sqrt(3x-5)=x-9+9-7x}}} Get all the non-radical terms to one side



{{{-4*sqrt(x-1)*sqrt(3x-5)=-6x}}} Combine like terms.



{{{-4*sqrt((x-1)(3x-5))=-6x}}} Combine the roots.



{{{sqrt((x-1)(3x-5))=(-6x)/(-4)}}} Divide both sides by -4.



{{{sqrt((x-1)(3x-5))=(3/2)x}}} Reduce



{{{(sqrt((x-1)(3x-5)))^2=((3/2)x)^2}}} Square both sides



{{{(x-1)(3x-5)=((3/2)x)^2}}} Square the square root



{{{(x-1)(3x-5)=(9/4)x^2}}} Square the right side



{{{4(x-1)(3x-5)=9x^2}}} Multiply both sides by 4.



{{{4(3x^2-8x+5)=9x^2}}} FOIL



{{{12x^2-32x+20=9x^2}}} Distribute



{{{12x^2-32x+20-9x^2=0}}} Subtract {{9x^2}}} from both sides.



{{{3x^2-32x+20=0}}} Combine like terms.



Notice that the quadratic {{{3x^2-32x+20}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-32}}}, and {{{C=20}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-32) +- sqrt( (-32)^2-4(3)(20) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-32}}}, and {{{C=20}}}



{{{x = (32 +- sqrt( (-32)^2-4(3)(20) ))/(2(3))}}} Negate {{{-32}}} to get {{{32}}}. 



{{{x = (32 +- sqrt( 1024-4(3)(20) ))/(2(3))}}} Square {{{-32}}} to get {{{1024}}}. 



{{{x = (32 +- sqrt( 1024-240 ))/(2(3))}}} Multiply {{{4(3)(20)}}} to get {{{240}}}



{{{x = (32 +- sqrt( 784 ))/(2(3))}}} Subtract {{{240}}} from {{{1024}}} to get {{{784}}}



{{{x = (32 +- sqrt( 784 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (32 +- 28)/(6)}}} Take the square root of {{{784}}} to get {{{28}}}. 



{{{x = (32 + 28)/(6)}}} or {{{x = (32 - 28)/(6)}}} Break up the expression. 



{{{x = (60)/(6)}}} or {{{x =  (4)/(6)}}} Combine like terms. 



{{{x = 10}}} or {{{x = 2/3}}} Simplify. 



So the <i>possible</i> solutions are {{{x = 10}}} or {{{x = 2/3}}} 

  

However, if you plug in {{{x = 2/3}}} into the original equation, you'll get a contradiction.


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Answer:



So the only solution is {{{x = 10}}}