Answer:
1) Methods:
- Quadratic formula.
- Factorization.
- Completing the square.
2) If the determinant is less than zero ([tex]D<0[/tex]) then there are two roots that are complex conjugates.
Step-by-step explanation:
Methods:
- Quadratic formula
Given the quadratic equation in Standard form [tex]ax^2+bx+c=0[/tex], you can solve it with the quadratic formula:
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
- Factorization
You must find two expression that when you multply them, you get the original quadratic equation. For example:
[tex]x^2+6x+8=0[/tex]
Find two number whose sum is 6 and whose product is 8. These are 2 and 4. Then:
[tex](x+2)(x+4)=0[/tex]
When you make the multiplication indicated in [tex](x+2)(x+4)=0[/tex], you obtain [tex]x^2+6x+8=0[/tex]
- Completing the square
Given the quadratic equation in Standard form [tex]ax^2+bx+c=0[/tex],, you must turn it into:
[tex]a(x+d)^2+e=0[/tex]
Where:
[tex]d=\frac{b}{2a}\\\\e=c-\frac{b^2}{4a}[/tex]
Once you get that form, you must solve for x.
You can predict if the quadratic function will have a complex solution with the determinant:
[tex]D=b^2-4ac[/tex]
If [tex]D<0[/tex] then there are two roots that are complex conjugates.
