Question

medal and fan
Determine the number and type of complex solutions and possible real solutions for each of the following equations.

1) 2x2 + 5x + 3 = 0

2) 4x3 – 12x + 9 = 0

3) 2x4 + x2 – x + 6 = 0

Answer 1

First, make sure your function is in quadratic form: \(ax^2+bx+c=0\)

From that you use the discriminant to determine real or complex roots.
Discriminant : \(b^2-4ac\)

\(b^2 - 4ac > 0 \) : 2 real roots that are unequal
\(b^2-4ac < 0 \) : 2 unreal (complex) roots that are unequal.
\(b^2-4ac = 0\) : 2 real equal roots.

First identify what your values of \(a~,~b~,~c\) are and then use the discriminant to find your real/complex roots :)

Answer 2

You can also apply Descartes' Rule of Signs here.

\(3)~ 2x^4 + x^2 – x + 6 = 0 \)
\[~~~~\color{red}{+}2x^4 \color{red}{+}x^2\color{blue}{-}x\color{red}{+}6=0\] Here, you can see that the sign changes twice, once between \(2x^4\) and \(x^2\) and the other between \(x^2\) and \(x\) and \(6\).

- 2 sign changes = 2 possible \(positive\) changes for the function

Answer 3

Now analyze the function by plugging in negative values into the x. \[f(-x) = 2(-x)^4 +(-x)^2-(-x)+6=0\] By simplifying this expression we will be able to tell how many complex solutions we have in our function. \[f(-x) = 2x^4 \color{red}{+}x^2 \color{red}{+}x\color{red}{+}6 = 0\] Since the sign \(did ~~not\) change, there \(does ~~ not\) seem to be any complex roots.

Answer 4

\(\therefore\) There are \(2\) real roots and \(0\) complex roots.

Answer 5

Srry, in the first post I made a typo. The sign change happens between \(x^2\) and \(x\) and \(x\) and \(6\).