1.

In order to have two identical roots `b^2-4ac` must be equal to 0, where a=k+4, b=k+1 and c=1.

`(k+1)^2-(4)(k+4)=0`

`k^2+2k+1-4k-16=0`

`k^2-2k-15=0`

Solve for k using the quadratic formula:

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

t where a=1, b=-2 and c=-15.

`x=(2+-sqrt((-2)^2-4(1)(-15)))/((2)(1))`

`x=(2+-sqrt(4+60))/((2))=(2+-sqrt64)/2`

`x=(2+8)/2=10/2=5`

`x=(2-8)/2=-6/2=-3`

Verification for k=5

`9x^2+6x+1=0`

`x=(-6+-sqrt(36-4(9)))/18=-1/3`

Verification for k=-3

`x^2-2x+1=0`

`x=(2+-sqrt(4-4))/2=1`

Therefore, k=5 and k=-3 are values that make one equal root.

2.

`f=s^2` (1)

`f+5s=66` (2)

Substitute (1) in (2) and write in the form ax^2+bx+c=0

`s^2+5s-66=0`

Factor to solve:

`(s-6)(s+11)=0`

s=6 and s=11

**If the son is 6 years old the father is 36 years old**. A
less likely solution would be the son being 11 years old and the father 121
years old.

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