Question

How do you solve `c^ { 2} - 30c - 216= 0`?

Answer 1

`c=36 or -6`

Explanation:

`c^2-30c-216=0`

Factor,

`(c-36)(c+6)=0`

Solve,

`c=36 or -6`

Answer 2

The solution set is {-6, 36}

Explanation:

The equation is a quadradic equation with a = 1, b = -30, and c = -216 (note this c is not the same c as the unknown quantity.)

One method of solving for c in the equation is to complete the square. This method only works if a = 1 or on an equation that you have done some manipulation on in order to make a = 1 (usually dividing both sides of the equation by a).

The complete the square method of solving a quadratic equation for its unknown is to add `(b/2)^2` to form a perfect square, then subtract the same quantity in order to preserve the equation.

In our case, `(b/2)^2` = `(-30/2)^2` = 225.

Starting with the original equation:
`c^2−30c−216=0`

Adding, then adding, then subtracting `(b/2)^2 = 225`:
`c^2−30c + 225 - 225 −216=0`
`c^2−30c + 225 - 441=0`

Then factoring our newly formed perfect square:
`(c^2−30c + 225) - 441=0`
`(c - 15)^2 - 441=0`

Then continue to solve for c:
`(c - 15)^2=441`
`sqrt((c - 15)^2)=sqrt(441)`
`|c - 15|` = 21
`c - 15 = +- 21`
`c= 15 +- 21`

Which implies that `c = 15 - 21 = -6` or `c = 15 + 21 = 36`.