To solve the equation 0=-3x^2+18x+27, we can start by factoring out -3 from the equation to get:
0 = -3(x^2-6x-9)
Next, we can use the quadratic formula to find the roots of the expression inside the parentheses:
x = (6 ± √(6^2 - 4(-9)))/2
x = (6 ± √72)/2
x = 3 ± 3√2
Therefore, the equation has two solutions, answer choice (B).
0=-3x^2+18x+27
A) one solution
B) two solutions
C) no solutions
D) infinitely many solutions
3 answers
Since it only wants to know about the nature of the solutions, of
-3x^2+18x+27 = 0 or
x^2 - 6x - 9 = 0
b^2 - 4ac
= 36 - 4(1)(-9) > 0
so we have 2 real solutions.
-3x^2+18x+27 = 0 or
x^2 - 6x - 9 = 0
b^2 - 4ac
= 36 - 4(1)(-9) > 0
so we have 2 real solutions.
If a and c have opposite signs, the discriminant is positive -- so 2 real roots