Answer:
1. x₁ = 6, x₂ = -6
2. x₁ = 3, x₂ = -3
Step-by-step explanation:
Given functions:
[tex]1)\ f(x) = -2(x-6)(x+6)[/tex]
[tex]2)\ f(x)=(x-3)(3x+9)[/tex]
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Zero Product Property: If m • n = 0, then m = 0 or n = 0.
Standard Form of a Quadratic: ax² + bx + c = 0.
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1. f(x) = -2(x - 6)(x + 6)
Step 1: Set the function to zero.
[tex]\implies 0 = -2(x-6)(x+6)[/tex]
Step 2: Divide both sides of equation by [tex]-2[/tex].
[tex]\implies \dfrac{0}{-2} = \dfrac{-2(x-6)(x+6)}{-2}[/tex]
[tex]\implies 0=(x-6)(x+6)[/tex]
Step 3: Apply the Zero Product Property.
[tex]x_1 \implies x-6=0[/tex]
[tex]x_2 \implies x+6=0[/tex]
Step 4: Solve for x in both equations.
[tex]x-6+6=0+6 \implies \boxed{x_1 = 6}[/tex]
[tex]x+6-6=0-6 \implies \boxed{x_1 = -6}[/tex]
The zeros (x-intercepts) of this function are: [tex]x_1=6,\ x_2=-6[/tex].
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2. f(x) = (x - 3)(3x + 9)
Step 1: Set the function to zero.
[tex]\implies 0 = (x - 3)(3x + 9)[/tex]
Step 2: Apply the Zero Product Property.
[tex]x_1 \implies x-3=0[/tex]
[tex]x_2 \implies 3x + 9=0[/tex]
Step 3: Solve for x in both equations.
[tex]x-3+3=0+3 \implies \boxed{x_1 = 3}[/tex]
[tex]3x=-9 \implies \dfrac{3x}{3}=\dfrac{-9}{3} \implies \boxed{x_1 = -3}[/tex]
The zeros (x-intercepts) of this function are: [tex]x_1=3,\ x_2=-3[/tex].
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