We know the quadratic equation is an equation of one variable with highest degree term 2. The quadratic with fractions means that the quadratic equation with fractions as their coefficient or
each term will be a fraction. Though there are different methods to solve a quadratic equation, we here stick on to the formula method.

If ax 2 + bx + c = 0, the solution of x is given by x = `[-b +- sqrt [b ^2 ** 4ac]] / [2a]` .

Now let us see few problems on this topic.

Example Problems on Solving Quadractic Equations with Fractions.

Ex 1: Solve: `1 /( x ** 2) + 2 /( x **1) = 6 / x` .

Soln: Given: `1 /( x ** 2) + 2 / (x ** 1) = 6 / x`

This can be simplified further as follows:

`[(x ** 1) + 2 (x ** 2) ]/ [(x ** 2) (x ** 1)]` = `6 / x`

`=>` `[3x ** 5] /[ x ^2 ** 3x + 2] = 6 / x`

= x (3x – 5) = 6 (x 2 – 3x + 2)

= 3x ^2 – 5x = 6x ^2 – 18x + 12.

Therefore 6x 2 – 18x + 12 – 3x 2 + 5x = 0

= 3x 2 – 13x + 12 = 0

Here a = 3, b = - 13, c = 12

Therefore x =`[ ** (**13) +- [sqrt [(**13) ^2 ** 4 (3) (12)]]] / [2 (3)]`

= `[13 +- [sqrt [169 ** 144]]] / 6`

= `[13 +- sqrt 25] / 6`

= `[13 +- 5] / 6`

Therefore x = `[13 + 5] / 6` , x = `[13 ** 5] / 6`

x = 3 or x = `4 / 3`

Ex 2: Solve: `(x + 3) / (x ** 2) ** (1 ** x) / x = 17 / 4`

Soln: Given:`( x + 3) / (x** 2) ** (1 ** x) / x = 17 / 4`

`[x (x + 3) ** (x ** 2) (1 ** x)] /[ x (x ** 2)] = 17 / 4`

`[x ^2 + 3x + x ^2 ** x + 2 ** 2x] / [x ^2 ** 2x] = 17 / 4`

`=>` `4 (2x ^2 + 2) = 17 (x ^2 ** 2x)`

`rArr` 8x 2 + 8 = 17 x 2 – 34x `=>` 17 x 2 – 8x 2 – 34 x – 8 = 0

`rArr` 9x 2 – 34 x – 8 = 0

`rArr` 9x 2 – 36x + 2x – 8 = 0

`=>` 9x (x – 4) + 2 (x – 4) = 0

`=>` (x – 4) (9x + 2) = 0

`=>` x = 4 or x = `- 2/9` .

Practice Problems on Solving Quadratic Equations with Fractions.

1. Solve the following :

(i) `[ x + 3] /[ x + 2 ]= [3x ** 7] / [2x ** 3]`

[Ans: x = -1 or x = 5]

2. `[2x] / [x ** 4] + [2x ** 5] / [x ** 3] = 25 / 3`

[Ans: x = 6 or x = `40 / 13` ]