Question

F(x)=x^2-3x+18 how many distinct real number zeros does f have?

Answer 1

Answer:

None of the distinct real number zeros the function have.

Step-by-step explanation:

Considering the function

$f(x)=x^2-3x+18$

The zeroes of a function are those values that touches the x-axis. In order to find those values we must

Set  $f(x) = 0$ in order to find those values

$0=x^2-3x+18$

$\mathrm{Switch\:sides}$

$x^2-3x+18=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-3,\:c=18:\quad x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$x=\frac{-\left(-3\right)+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$x=\frac{3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

As

$3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}=3+\sqrt{63}i$

so

$x=\frac{3+\sqrt{63}i}{2\cdot \:1}$

$x=\frac{3+3\sqrt{7}i}{2}$

$x=\frac{3}{2}+\frac{3\sqrt{7}}{2}i$

Similarly,

$x=\frac{-\left(-3\right)-\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}:\quad \frac{3}{2}-i\frac{3\sqrt{7}}{2}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{3}{2}+i\frac{3\sqrt{7}}{2},\:x=\frac{3}{2}-i\frac{3\sqrt{7}}{2}$

BUT, NONE OF THEM ARE REAL NUMBER ZEROS.

Therefore, none of the distinct real number zeros the function have.