Question

select the correct answer.
which function has a zero with a multiplicity of 2?
\\( f(x) = x^2 + 3x + 9 \\)
\\( f(x) = 3x^2 + 24x + 48 \\)
\\( f(x) = -3x^2 + 12x + 12 \\)
\\( f(x) = x^2 - 2x - 1 \\)

Explanation:

Step1: Recall multiplicity condition

A quadratic has a zero with multiplicity 2 if its discriminant $\Delta = b^2-4ac = 0$.

Step2: Calculate $\Delta$ for Option 1

For $f(x)=x^2+3x+9$, $a=1,b=3,c=9$:
$\Delta = 3^2 - 4(1)(9) = 9 - 36 = -27
eq 0$

Step3: Calculate $\Delta$ for Option 2

For $f(x)=3x^2+24x+48$, $a=3,b=24,c=48$:
$\Delta = 24^2 - 4(3)(48) = 576 - 576 = 0$

Step4: Verify other options (optional)

For $f(x)=-3x^2+12x+12$, $a=-3,b=12,c=12$:
$\Delta = 12^2 - 4(-3)(12) = 144 + 144 = 288
eq 0$
For $f(x)=x^2-2x-1$, $a=1,b=-2,c=-1$:
$\Delta = (-2)^2 - 4(1)(-1) = 4 + 4 = 8
eq 0$

Answer:

B. $f(x) = 3x^2 + 24x + 48$