Question

How do you solve `0=p^2-p-11`?

Answer 1

Use the quadratic formula to find `p = (1+-3sqrt(5))/2`

Explanation:

If `ap^2+bp+c = 0` then

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

We have `a=1`, `b=-1`, `c=-11`

So:

`p = (1+-sqrt(1+44))/2 = (1+-sqrt(45))/2`

`= (1+-sqrt(3^2*5))/2 = (1+-3sqrt(5))/2`

Hence solutions are
`p = (1+-3sqrt(5))/2`

Answer 2

`p=(1+3sqrt5)/2, (1-3sqrt5)/2`

Explanation:

`p^2−2p(1/2)+(1/4)-(11+1/4)=0 or, (p−1/2)^2=45/4=5(3/2)^2 or, [option 1] p−1/2 = 3sqrt5/2 or, p=(1+3sqrt5)/2 or, [option 2] p−1/2 = -3sqrt5/2 or, p=(1-3sqrt5)/2`