Respuesta :
Answer:
The solution are imaginary numbers (complex numbers)
[tex]\begin{gathered} x=\frac{10+i\sqrt[]{20}}{2}=5+i\frac{\sqrt[]{20}}{2}=5+2.24i \\ \text{and} \\ x=\frac{10-i\sqrt[]{20}}{2}=5-i\frac{\sqrt[]{20}}{2}=5-2.24i \end{gathered}[/tex]Explanation:
Given the quadratic function;
[tex]f(x)=x^2-10x+30[/tex]The zeros of the function are at f(x)=0;
[tex]\begin{gathered} f(x)=x^2-10x+30=0 \\ x^2-10x+30=0 \end{gathered}[/tex]Solving using the quadratic formula;
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}_{}}{2a}[/tex]where;
a = 1
b = -10
c = 30
Substituting the values, we have;
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}_{}}{2a} \\ x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(1)(30)}_{}}{2(1)} \\ x=\frac{10\pm\sqrt[]{100^{}-120}_{}}{2} \\ x=\frac{10\pm\sqrt[]{-20}_{}}{2} \\ x=\frac{10\pm\sqrt[]{(-1)20}_{}}{2} \\ x=\frac{10\pm\sqrt[]{(i^2)20}_{}}{2} \\ x=\frac{10\pm i\sqrt[]{20}_{}}{2} \end{gathered}[/tex]Recall that;
[tex]-1=i^2[/tex]Solving further we have;
[tex]\begin{gathered} x=\frac{10+i\sqrt[]{20}}{2}=5+i\frac{\sqrt[]{20}}{2}=5+2.24i \\ \text{and} \\ x=\frac{10-i\sqrt[]{20}}{2}=5-i\frac{\sqrt[]{20}}{2}=5-2.24i \end{gathered}[/tex]Therefore, the solution are imaginary numbers (complex numbers)
