Question

if $a,b,c,d\in R$, 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

Consider the given equation,

$(x^2+ax-3b)(x^2-cx+b)(x^2-dx+2b)=0$

If, $a,b,c,d\in R$

Then,

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

Comparing that,

$A{x}^2+Bx+C=0$

We know that,

$D=B^2-4AC$

$D=a^2-4\left(1\right)(-3b)$

$D=a^2+12b$

Now $D=0$ then,

$a^2+12b=0$

$b=-\frac{a^2}{12}<0\left(\text{Imaginery}\text{roots}\right)$

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

Comparing that,

$A{x}^2+Bx+C=0$

We know that,

$D=B^2-4AC$

$D=c^2-4\left(1\right)\left(b\right)$

$D=c^2-4b$

Now,$D=0$ then,

$c^2-4b=0$

$c^2=4b$

$c=2\sqrt{b}>0\left(\text{Real}\text{roots}\right)$

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

Comparing that,

$A{x}^2+Bx+C=0$

We know that,

$D=B^2-4AC$

$D={(-d)}^2-4\left(1\right)\left(2b\right)$

$D=d^2+8b$=

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

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

Hence, this is the answer.