Answer:
The two zeros of the quadratic function are:
[tex]x=\frac{2}{3},\:x=-\frac{2}{3}[/tex]
Step-by-step explanation:
Given the expression
[tex]9x^2-4[/tex]
In order to determine the zeros of the quadratic function, we get the equation
[tex]9x^2-4=0[/tex]
Add 4 to both sides
[tex]9x^2-4+4=0+4[/tex]
Simplify
[tex]9x^2=4[/tex]
Divide both sides by 9
[tex]\frac{9x^2}{9}=\frac{4}{9}[/tex]
Simplify
[tex]x^2=\frac{4}{9}[/tex]
For x² = f(a) the solutions are: [tex]x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{\frac{4}{9}},\:x=-\sqrt{\frac{4}{9}}[/tex]
solving
[tex]x=\sqrt{\frac{4}{9}}[/tex]
[tex]=\frac{\sqrt{4}}{\sqrt{9}}[/tex]
[tex]=\frac{2}{3}[/tex]
also solving
[tex]x=-\sqrt{\frac{4}{9}}[/tex]
[tex]=-\frac{\sqrt{4}}{\sqrt{9}}[/tex]
[tex]=-\frac{2}{3}[/tex]
Therefore, the two zeros of the quadratic function are:
[tex]x=\frac{2}{3},\:x=-\frac{2}{3}[/tex]
