Question

If a , b, c d $ϵR$, 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

Answer: (B)

at least 2 real roots

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

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

$D=(a{)}^2-4(1\left)\right(-3b)=a^2+12b$

For $x^2-cx+b=0D=(-c{)}^2-4(1\left)\right(b)=c^2-4b$

For $x^2-dx+2b=0D=(-d{)}^2+4(1\left)\right(26)=d^2+8b$

Let us assume that $x^2+ax-3b$ has imaginary $y$ axis

then $a^2+12b<012b<-a^{2}b<\frac{-a^2}{12}$

$\therefore b$ is (-ve) then $-4b$ willbe (+ve)

$c^{2}is(+ve)$

So, $c^2-4b$ will be $(+ve)$

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

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

Has $2$ real roots.