Question

The function f is defined by f(x) = x2 - 6x +21.

What are the solutions of f(x) = 0?
Show your work on the scratchpad.

Answer 1

The solutions are:

`x =3+2√(3)i\\\\\:x=3-2√(3)i`

Solution:

`f(x) = x^2 - 6x + 21`

We have to find solutions when f(x) = 0

`x^2 - 6x + 21 = 0`

Solve by quadratic formula

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=(-b\pm √(b^2-4ac))/(2a)\\\\\mathrm{For\:}\quad a=1,\:b=-6,\:c=21\\\\x=(-\left(-6\right)\pm √(\left(-6\right)^2-4\cdot \:1\cdot \:21))/(2\cdot \:1)`

`x = (6 \pm √(\left(-6\right)^2-4\cdot \:1\cdot \:21))/(2\cdot \:1)\\\\Simplify\\\\x = (6 \pm √(36 - 84))/(2)\\\\x = (6 \pm √(-48))/(2)\\\\x = (6 \pm i √(48))/(2)\\\\simplify\\\\x = (6 \pm 4√(3)i)/(2)\\`

`x = 3 \pm 2 √(3) i`

We have two solutions:

`x = 3 + 2 √(3) i\\\\x = 3 - 2 √(3) i`

Thus the solutions are:

`x =3+2√(3)i\\\\\:x=3-2√(3)i`