# Complex numb help (1 Viewer)

#### amdspotter

##### Member

I understand part b and c but i dont get part a.

#### CM_Tutor

##### Moderator
Moderator
$\bg_white P(x)$ is a quadratic equation in $\bg_white x^2$, so the solutions of $\bg_white P(x) = 0$ are

$\bg_white x^2 = \cfrac{-A \pm \sqrt{A^2 - 4(1)(B)}}{2}$

$\bg_white \text{and since}\ \sqrt{A^2 - 4B} < A\ \implies\ x^2 < 0$

there are no real solutions.

#### icycledough

##### Well-Known Member
I'm not sure if you would be required to justify it mathematically it, or if you could explain it in writing. But essentially, x^4 will always be positive, Ax^2 will always be positive (as A is strictly positive) and B must be positive. Thus, adding 3 positive numbers cannot result in a root for the equation.

#### Drongoski

##### Well-Known Member
I'm not sure if you would be required to justify it mathematically it, or if you could explain it in writing. But essentially, x^4 will always be positive, Ax^2 will always be positive (as A is strictly positive) and B must be positive. Thus, adding 3 positive numbers cannot result in a root for the equation.
That's essentially what I said before. The simplest explanation.

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