Question

If $a,b,c,r\in R$, then the equation $(x^2+ax-3b)(x^2-cx+b)(x^2-dx+2b)=0$ has

• A. $6$ real roots
• B. at least $2$ real roots
• C. $4$ real roots
• D. $3$ real roots

Answer 1

The correct option is B at least $2$ real roots

Given equation : $[x^2+ax-3b][x^2-cx+b][x^2-dx+2b]=0$

For $x^2+ax-3b=0$

Discriminant, $D=a^2+12b$

For $x^2-cx+b=0$

Discriminant, $D=c^2-4b$

For $x^2-dx+2b=0$

Discriminant, $D=d^2+8b$

Let us assume that $x^2+ax-3b$ has imaginary roots, then,

$a^2+12b<0$

$b<-a^2/12$

this implies b is negative, then $-4b$ will be positive

so,

$c^2-4b>0$

$x^2-cx+b=0$ has real roots.

Therefore, $(x^2+ax-3b)(x^2-cx+b)(x^2-dx-2b)=0$ has at least 2 real roots.