Answer:
None of the distinct real number zeros the function have.
Step-by-step explanation:
Considering the function
[tex]f(x)=x^2-3x+18[/tex]
The zeroes of a function are those values that touches the x-axis. In order to find those values we must
Set [tex]f(x) = 0[/tex] in order to find those values
[tex]0=x^2-3x+18[/tex]
[tex]\mathrm{Switch\:sides}[/tex]
[tex]x^2-3x+18=0[/tex]
[tex]\mathrm{Solve\:with\:the\:quadratic\:formula}[/tex]
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=1,\:b=-3,\:c=18:\quad x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}[/tex]
[tex]x=\frac{-\left(-3\right)+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}[/tex]
[tex]x=\frac{3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}[/tex]
As
[tex]3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}=3+\sqrt{63}i[/tex]
so
[tex]x=\frac{3+\sqrt{63}i}{2\cdot \:1}[/tex]
[tex]x=\frac{3+3\sqrt{7}i}{2}[/tex]
[tex]x=\frac{3}{2}+\frac{3\sqrt{7}}{2}i[/tex]
Similarly,
[tex]x=\frac{-\left(-3\right)-\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}:\quad \frac{3}{2}-i\frac{3\sqrt{7}}{2}[/tex]
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=\frac{3}{2}+i\frac{3\sqrt{7}}{2},\:x=\frac{3}{2}-i\frac{3\sqrt{7}}{2}[/tex]
BUT, NONE OF THEM ARE REAL NUMBER ZEROS.
Therefore, none of the distinct real number zeros the function have.
