Question

How do you solve `x^2 + 9x + 9 = 0` using the quadratic formula?

Answer 1

Root1=`-1.146` and Root2 = `-7.854`

Explanation:

This is quadratic equation where `a=1 ;b=9; c=9; b^2-4ac > 0` So roots are real. Roots are `-b/(2*a) +- sqrt (b^2-4ac)/(2a)` `:.`Root1 `=-9/2+sqrt45/2 = -1.146` and Root2`=-9/2-sqrt45/2 = -7.854`[Ans]

Answer 2

`(-9 +- 3sqrt5)/2`

Explanation:

`y = x^2 + 9x + 9 = 0`
Use the improved quadratic formula (Socratic Search)
`D = d^2 = b^2 - 4ac = 81 - 36 = 45`--> `d = +- 3sqrt5`
There are 2 real roots:

`x = -b/(2a) +- d/(2a) = -9/2 +- 3sqrt5/2 = (-9 +- 3sqrt5)/2`