Answer: [tex]x=\frac{7\pm 3i\sqrt{3}}{2}[/tex]
Explanation:
Since, given quadratic function, [tex](x-5)^2 + 3(x-5) + 9 = 0[/tex] -----(1)
let us consider,[tex]x-5=p[/tex] -----(2)
Put this value in equation (1),
We get, [tex]p^2+3 p+9=0[/tex], which is also a quadratic equation.
Since, quadratic formula for finding the root of quadratic equation of type [tex]ax^2+bx+c=0[/tex] is, [tex]x=\frac{-b\pm\sqrt{b^2-4ac}} {2a}[/tex]
Here, a=1, b=3 and c=9, so, [tex]p=\frac{-3\pm\sqrt{3^2-4\times1 \times9}} {2\times1}[/tex]
Thus, [tex]p=\frac{-3\pm3\sqrt{-3}} {2}[/tex] ⇒ [tex]p=\frac{-3\pm3i\sqrt{3}} {2}[/tex] (because √-1=i)
Now, from equation(2) [tex]x-5=\frac{-3\pm3i\sqrt{3}} {2}[/tex]
⇒[tex]x=\frac{-3\pm3i\sqrt{3}} {2}+5=\frac{7\pm 3i\sqrt{3}}{2}[/tex]
⇒[tex]x=\frac{7\pm 3i\sqrt{3}}{2}[/tex]
