Question

How do you solve `27x ^ { 2} - 55x + 2= 0`?

Answer 1

`x = 2, 1/27`

Explanation:

We can use two methods here; i) Factorisation Method, or ii)Quadratic Formula

I'm Going with the Factorisation method.

So,

`color(white)(xx)27x^2 - 55x + 2 = 0`

`rArr 27x^2 - (54 + 1)x + 2 = 0` [Wrote `55` as `(54 + 1)`]

`rArr 27x^2 - 54x - x +2 = 0` [Distributive Property]

`rArr 27x(x - 2) - 1(x - 2) = 0` [Grouping the like terms]

`rArr (x - 2)(27x - 1) = 0` [Grouped Again]

Now, We know, If the product of two real numbers is zero, then one of them must be zero. (Both can be zero too.)

So, `x - 2 = 0 rArr x = 2`

Or, `27x - 1 = 0 rArr 27x = 1 rArr x = 1/27`

So, We have, `x = 2, 1/27`

If you want to use the Quadratic Formula, then you have to check first that the equation is in the form `ax^2 + bx + c = 0` (General) form or not.

So, Here the equation is fine.

Now using the Quadratic Formula.

It states that, A quadratic equation of form `ax^2 + bx + c = 0` has roots :-

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

So, We should check the Discriminant `(b^2 - 4ac)` first.

Here, `D = b^2 -4ac = (-55)^2 - 4 * 27 * 2 = 3025 - 216 = 2809 gt 0`

So, We should get two roots which are distinct and real.

So, `x = (-b + sqrt(D))/(2a) = (- (-55) + sqrt(2809))/(2 * 27) = (55 + 53)/(54) = 108/54 = 2`

And, `x =(-b - sqrt(b^2 - 4ac))/(2a) = (-(-55) - sqrt(2809))/(2 * 27) = (55 - 53)/54 = 2/54 = 1/27`

So, We get, `x = 2, 1/27`

Hence Explained.