Question

Which of the following statement is true about the equation $x^4+2x^2-8x+3=0?$

• A. There are $2$ positive real roots, $2$ imaginary roots
• B. There are $4$ imaginary roots
• C. There are $2$ negative real roots
• D. There are $1$ positive real root, $1$ negative real root, $2$ imaginary roots

Answer 1

The correct option is A There are $2$ positive real roots, $2$ imaginary roots

Let $f\left(x\right)=x^4+2x^2-8x+3$.
According to the Fundamental Theorem of Algebra, a $n$ degree polynomial equation should have $n$ roots.
Thus, $f\left(x\right)$ should also have $4$ roots (real or imaginary).
Using Descartes Rule of Signs,
The number of sign changes in $f\left(x\right)$ is $2$
$\Rightarrow$ Maximum number of positive real roots $=2$.
$\Rightarrow$ Possible number of positive real roots $=2$ or $0$.

$\&f(-x)=(-x{)}^4+2(-x{)}^2-8(-x)+3$
$\Rightarrow f(-x)=x^4+2x^2+8x+3$
Thus, the number of sign changes in $f\left(x\right)$ is $0$
$\Rightarrow$ Maximum number of negative real roots $=0$
$\Rightarrow$ Possible number of negative real roots $=0$

From the above observations, the different possibilities are given below.

Now, let's try to find the value of $f\left(x\right)$ at different points
$f\left(0\right)=(0{)}^4+2(0{)}^2-8\left(0\right)+3=3$
$f\left(1\right)=(1{)}^4+2(1{)}^2-8\left(1\right)+3=-2$
Here, $f\left(0\right)>0$ and $f\left(1\right)<0$
$\Rightarrow f\left(x\right)$ cuts $x$ axis at least once between $0\text{and}1.$
$\Rightarrow$ The given equation has at least one real root.
$\Rightarrow$ Possibility $1$ is valid.