Descares Theorem says that the number of roots of a polynomial is equal to its highest exponent, and the roots could be real and/or imaginary (complex).

However is their is one complex root (a + i.b), it's conjugate is also
a root (a-i.b).

In the equation x⁴ - 4x³ - 2x² + 12x + 9 = 0, we have already 3 roots, then the 4rth cannot be a complex one since we don't have a fifth for its conjugate.

3 0

3 years ago

Alexis Brian and Carli collect graphic novels, Brian has 5 more graphic novels then Alexis and Carli has twice as many as Alexis

kkurt [141]

According to Google, a Variable is <span>any factor, trait, or condition that can exist in differing amounts or types. If you type in variable in the Google search bar, it will give you the definition in a little box at the very top of the page :).
</span>

4 0

3 years ago

Find all complex solutions of 5x^2-6x+2=0.

gregori [183]

We use quadratic formula for that one
remember that √-1=i

so

for
ax^2+bx+c=0
$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$
so
given
5x^2-6x+2=0
a=5
b=-6
c=2

$x=\frac{-(-6)+/-\sqrt{(-6)^2-4(5)(2)}}{2(5)}$
$x=\frac{6+/-\sqrt{36-40}}{10}$
$x=\frac{6+/-\sqrt{-4}}{10}$
$x=\frac{6+/-2i}{10}$

$x=\frac{3}{5}+\frac{i}{5}, \frac{3}{5}-\frac{i}{5}$

8 0

3 years ago