Question

The polynomial $(a{x}^2+bx+c)(a{x}^2-dx-c)$, $ac\ne 0$ has

• A. Four real roots
• B. At least two real roots
• C. At most two real roots
• D. No real roots

Answer 1

The correct option is B

At least two real roots

Explanation for the correct option:

Find the number of real roots of the given polynomial

A polynomial $(a{x}^2+bx+c)(a{x}^2-dx-c)$ is given where $a·c\ne 0$.

Step 1 : Find the nature of roots when $a·c<0$.

Find the discriminant $D_1$ of $\left(a{x}^2+bx+c\right)$.

$D_1=b^2-4ac$

Since, $a·c<0$. so, $D_1>0$.

Therefore, when $a·c<0$ polynomial $\left(a{x}^2+bx+c\right)$ has two real roots.

Find the discriminant $D_2$ of $\left(a{x}^2-dx-c\right)$.

$D_2=d^2+4ac$

Since, $a·c<0$. so, $D_2$ may be positive or negative.

Therefore, when $a·c<0$ polynomial $(a{x}^2+bx+c)(a{x}^2-dx-c)$ has at least two real roots.

Step 2 : Find the nature of roots when $a·c>0$.

Find the discriminant $D_1$ of $\left(a{x}^2+bx+c\right)$.

$D_1=b^2-4ac$

Since, $a·c>0$. so, $D_1$ may be positive or negative.

Find the discriminant $D_2$ of $\left(a{x}^2-dx-c\right)$.

$D_2=d^2+4ac$

Since, $a·c>0$. so, $D_2>0$.

Therefore, the polynomial $\left(a{x}^2-dx-c\right)$ has two real roots.

Thus, the polynomial $(a{x}^2+bx+c)(a{x}^2-dx-c)$ has at least two real roots.

Hence, option $B$ is the correct answer.