How do you solve c^ { 2} - 30c - 216= 0c230c216=0?

2 Answers
May 14, 2018

c=36 or -6c=36or6

Explanation:

c^2-30c-216=0c230c216=0

Factor,

(c-36)(c+6)=0(c36)(c+6)=0

Solve,

c=36 or -6c=36or6

May 14, 2018

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(b2)2 to form a perfect square, then subtract the same quantity in order to preserve the equation.

In our case, (b/2)^2(b2)2 = (-30/2)^2(302)2 = 225.

Starting with the original equation:
c^2−30c−216=0c230c216=0

Adding, then adding, then subtracting (b/2)^2 = 225(b2)2=225:
c^2−30c + 225 - 225 −216=0c230c+225225216=0
c^2−30c + 225 - 441=0c230c+225441=0

Then factoring our newly formed perfect square:
(c^2−30c + 225) - 441=0(c230c+225)441=0
(c - 15)^2 - 441=0(c15)2441=0

Then continue to solve for c:
(c - 15)^2=441(c15)2=441
sqrt((c - 15)^2)=sqrt(441)(c15)2=441
|c - 15||c15| = 21
c - 15 = +- 21c15=±21
c= 15 +- 21c=15±21

Which implies that c = 15 - 21 = -6c=1521=6 or c = 15 + 21 = 36c=15+21=36.