Question 64609
QUESTION:


sqrt(2x-1) - sqrt(x+3) = 1


ANSWER:

sqrt(2x-1) - sqrt(x+3) = 1


Squaring on both sides,


(sqrt(2x-1) - sqrt(x+3))^2 = 1^2



(2x-1) + (x+3)- 2 sqrt[ (2x-1)(x+3) ] = 1



2x-1 + x+3 - 2 sqrt[ 2x*x + 2x* 3- 1 * x -1 * 3 ] = 1


2x + x - 1 + 3 - 2 sqrt[2x^2 + 6x - x - 3 ] = 1


3x + 2 - 2sqrt [2x^2 + 5x -3] = 1


3x + 2 - 1 = 2sqrt [2x^2 + 5x -3] 



3x + 1 = 2sqrt [2x^2 + 5x -3] 


Squaring again,


(3x + 1 )^2 = (2sqrt [2x^2 + 5x -3] )^2



9x^2 + 6x + 1 = 4  [2x^2 + 5x -3]


9x^2 + 6x + 1 = 4 * 2x^2 + 4 *5x - 4*3 



9x^2 + 6x + 1 = 8x^2 + 20x - 12


9x^2 -8x^2 + 6x  - 20x + 1 + 12 = 0


1x^2 - 14x + 13 = 0



x^2 - 14x + 13 = 0



This is a quadratic equation,


For solving this equation we have different methods - 1)Using quadratic formula 2) Splitting middle term.


Splitting middile term:

x^2 - 14x + 13 = 0


Here we have to find out two numbers whose sum is -14 and product is +13.
Such two numbers are -13 and -1


==> x^2 - 13x - 1x+ 13 = 0



==> ( x^2 - 13x )- (1x - 13 ) = 0


==> x (x -13) -1(x-13) = 0


Take out common terms.


==> (x-13)(x-1) = 0


==> either (x-13)= 0  or (x-1) = 0


==> x = 13   or  x = 1



So the solution is  x = 13   or  x = 1


2. Quadratic formula:


The general form of a quadratic equation is ax^2 + bx + c = 0 and its solution is given by the formula,


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ______________________(1)


Here we have  a = 1, b = -14 and  c = 13


Substitute these values in (1)



We have, {{{x = (-( -14 ) +- sqrt( (-14)^2-4*1*13 ))/(2*1) }}} 



==> {{{x = (14  +- sqrt( 196 -52))/ (2)}}} 


==> {{{x = (14  +- sqrt( 144))/ (2)}}} 


==> {{{x = (14  +- 12)/ (2)}}} 



==> x = (14 + 12 )/ 2  or  x = (14 - 12 )/ 2 



==> x = 26/2 or x = 2/2



==> x = 13 or x = 1



In both cases we can see that answers are same.



Hope you understood.


Regards.


Praseena.