Question

How do you solve `p^ { 2} - 12p - 73= 0`?

Answer 1

`p = 6+-sqrt(109)`

Explanation:

We can solve this by completing the square and using the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with `A=(p-6)` and `B=sqrt(109)` as follows:

`0 = p^2-12p-73`

`color(white)(0) = p^2-2p(6)+36 -36-73`

`color(white)(0) = p^2-2p(6)+36-109`

`color(white)(0) = (p-6)^2-(sqrt(109))^2`

`color(white)(0) = ((p-6)-sqrt(109))((p-6)+sqrt(109))`

`color(white)(0) = (p-6-sqrt(109))(p-6+sqrt(109))`

Hence:

`p = 6+-sqrt(109)`

Answer 2

`p=6+-sqrt109`

Explanation:

Use the quadratic formula. `x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)`

where `a = 1," "b = -12 and c = -73`

When you plug everything in, you get two answers:

`p=6+sqrt109` and `p=6-sqrt109`.