Form a quadratic equation with integer coefficients and one of its roots as 3 + 4*sqrt 5.

Let the equation be ax^2 + bx + c
=0

Now we have one of the roots as  3+ 4sqrt5, let the
other root of the equation be R.

So we have (x- 3 - 4*sqrt
5)(x- R) = ax^2 + bx + c

=> x^2 – x*(3 - 4sqrt 5) –
Rx + (3+4 sqrt 5)R = ax^2 + bx + c

=> x^2 – x*( 3-
4sqrt5 + R) + (3+4 sqrt 5)R = ax^2 + bx + c

Now if c is an
integer R has to be equal to 3- 4 sqrt 5 as (3+4 sqrt 5) (3-4 sqrt 5) = 9 – 16*5 = 9 –
80 = -71

Therefore (x- 3 + 4*sqrt 5)(x- R) = ax^2 + bx +
c

=> (x- 3 - 4*sqrt 5)(x- 3+ 4sqrt5) = ax^2 + bx +
c

=> x^2 + x( -3-4sqrt5 -3 + 4 sqrt5) + 9 – 80 =
ax^2 + bx + c

=> x^2  -6x -72 = ax^2 + bx +
c

=> a =1 , b = -6 and c=
-71

Therefore the equation is x^2 -6x -71 =
0