You can put this solution on YOUR website! Solve
75/(x+12) + 9/x = 3
[75x+9(x+12)]/x(x+12)=3 (finding the lcm and simplifying thus combining the sum of two fractions on the LHS to form a single fraction )
[75x+9(x+12)]=3x(x+12) (multiplying both the sides by x(x+12)
[75x+9x+108]= 3x^2+36
84x+108 =3x^2+36
0 = 3x^2-84x+36-108 (subtracting 84x and 108 from both the sides)
0 = 3x^2-84x-72
That is 3x^2-84x-72 = 0
(LHS=RHS is the same as RHS=LHS since equality is symmetric)
Dividing by 3 throught out
x^2 -28x -24 =0 ----(1) (zero divided by anything is zero)
x^2 -28x = +24
x^2 -2X(14)x + (14^2) = 24 +(14^2)
(adding (14^2) to both the sides inorder to make the LHS a perfect square)
(x-14)^2 = (24+196)
(x-14)^2 = 220
(x-14) = +or-(sqrt(220)
x = 14 +or-(sqrt(220)
Answer:x = [14 +(sqrt(220)]and x = [14 -(sqrt(220)]
Verification: Putting [14 +(sqrt(220)]in (1)
LHS = [14 +(sqrt(220)]^2 -28[14 +(sqrt(220)] -24
=(14+u)^2-28(14+u)-24 where u = +sqrt(220)
=[(14^2) +2X14Xu +u^2]-392-28u-24
(using (a+b)^2 = a^2 +2ab +b^2 where a=14 b =u)
=196 +28u +u^2-392-28u -24
=(196-392-24)+(28u-28u)+(220) (putting the value of u^2 which is 220)
= (196+220)-(392+24)+0
=416-416
=0
=RHS
Verification: Putting x=[14 -(sqrt(220)]in (1)
LHS = [14 -(sqrt(220)]^2 -28[14 -(sqrt(220)] -24
=(14-u)^2-28(14-u)-24 where u = +sqrt(220)
=[(14^2) +2X14X(-u) +u^2]-392+28u-24
(using (a+b)^2 = a^2 +2ab +b^2 where a=14 b =-u)
=196 -28u +u^2-392+28u -24
=(196-392-24)+(28u-28u)+(220) (putting the value of u^2 which is 220)
= (196+220)-(392+24)+0
=416-416
=0
=RHS
Note: Why have not used factorisation method for finding the vaues of x?
It is because in the equation x^2 -28x -24 =0 ----(1)
it is not possible to split the mid term -28x as the sum of two integer parts whose product is(x^2)X(-24) = -24x^2 as mid term negative and product negative means the two parts should be of opp signs and their addition amounts to giving the sign of the larger and subtracting the smaller from the larger and the factors of 24 cannot give a difference of 28 in terms of integers.
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