Question

If the equation $x^3-8x^2+cx+d=0\forall c,d\in R$ has one complex root and one positive root, then select the correct statement.

• A. $c,d>0$
• B. $c<0,d>0$
• C. $c<0,d<0$
• D. $c>0,d<0$

Answer 1

Answer: (C), (D)

$c>0,d<0$

Let $f\left(x\right)=x^3-8x^2+cx+d$
Now, one of the roots of the equation is complex.

From fundamental theorem of algebra: The complex roots of a polynomial equation with real coefficients, if exist, always occurs in conjugate pair.
$\Rightarrow$ The number of real roots will be $1.$
Now, $f\left(x\right)$ has one positive real root as well.
Thus, the number of sign changes in $f\left(x\right)$ should be $3$ or $1.$
$\text{Case}1.c,d>0$
In this case $f\left(x\right)=x^3-8x^2+cx+d$ has $2$ sign changes, which is not possible.
$\text{Case}2.c>0,d<0$
In this case $f\left(x\right)=x^3-8x^2+cx-|d|$ has 3 sign changes, which is possible.
$\text{Case}3.c<0,d>0$
In this case $f\left(x\right)=x^3-8x^2-|c|x+d$ has $2$ sign changes, which is not possible.
$\text{Case}4.c<0,d<0$
In this case $f\left(x\right)=x^3-8x^2-|c|x-|d|$ has $1$ sign changes, which is possible.

Thus, for $\text{Case}2.\text{and Case}3.$ the equation will have one positive real root.