Question

Solve for x : \[\dfrac{1}{{x - 3}} + \dfrac{2}{{x - 2}} = \dfrac{8}{x}\] ; \[x \ne 0,2,3\].

Answer 1

Hint: Given is an expression on solving will be a quadratic equation. We will solve that equation and will find two values of x such that the values are called roots of the equation. We will first take LCM on LHS and then on cross multiplying we will face a quadratic equation. Then we will use a quadratic formula to solve it. That’s it!

Complete step-by-step answer:
Given is,
\[\dfrac{1}{{x - 3}} + \dfrac{2}{{x - 2}} = \dfrac{8}{x}\]
Now we will take LCM on LHS,
\[ \Rightarrow \dfrac{{1\left( {x - 2} \right) + 2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} = \dfrac{8}{x}\]
On multiplying the brackets,
\[ \Rightarrow \dfrac{{x - 2 + 2x - 6}}{{{x^2} - 2x - 3x + 6}} = \dfrac{8}{x}\]
Rearranging the terms we get,
\[ \Rightarrow \dfrac{{3x - 8}}{{{x^2} - 5x + 6}} = \dfrac{8}{x}\]
Now again we will cross multiply the terms as,
\[ \Rightarrow \left( {3x - 8} \right)x = 8\left( {{x^2} - 5x + 6} \right)\]
Now we will multiply the terms inside the bracket with the term outside the bracket,
\[ \Rightarrow 3{x^2} - 8x = 8{x^2} - 40x + 48\]
Now we will take all the term son one side of the equation so that we can form the quadratic equation,
\[ \Rightarrow 8{x^2} - 3{x^2} - 40x + 8x + 48 = 0\]
Now performing the operations on similar terms,
\[ \Rightarrow 5{x^2} - 32x + 48 = 0\]
There is the equation of the form \[a{x^2} + bx + c = 0\] and we will solve this using the quadratic equation formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
So on comparing with the general equation and then putting the values in the quadratic formula we get,
\[ = \dfrac{{ - \left( { - 32} \right) \pm \sqrt {{{\left( { - 32} \right)}^2} - 4 \times 5 \times 48} }}{{2 \times 5}}\]
Now solving the root first,
\[ = \dfrac{{32 \pm \sqrt {1024 - 960} }}{{10}}\]
Perform the subtraction now,
\[ = \dfrac{{32 \pm \sqrt {64} }}{{10}}\]
Now we know that 64 is the perfect square of 8 so taking the root,
\[ = \dfrac{{32 \pm 8}}{{10}}\]
Now we will split the terms as,
\[ = \dfrac{{32 + 8}}{{10}}or\dfrac{{32 - 8}}{{10}}\]
On adding and subtracting,
\[ = \dfrac{{40}}{{10}}or\dfrac{{24}}{{10}}\]
On dividing by 10 we get,
\[x = 4orx = 2.4\]
Thus x has the two values mentioned above.

Note: Note that we cannot solve the equation above by trial-and-error method for different values of x. so we will solve the expression to a maximum simplified answer and then will turn towards the solution. Also note that to cross verify the values we can put them back in the given equation and then if the RHS is equal to the LHS then we are having the correct answer.
Also since the equation is quadratic the values of x should be two definitely.