Question

Let a and b be two distinct roots of the equation $x^3+3x^2-1=0$. The equation which has (ab) as its roots is equal

• A. $x^3-3x^2-1=0$
• B. $x^3-3x-1=0$
• C. $x^3+x^2-3x+1=0$
• D. $x^3+x^2+3x-1=0$

Answer 1

Answer: (B)

$x^3-3x-1=0$

Given:$x^3+3x^2-1=0$

Let $a,b$ and $c$ be the roots of the above equation

$\Rightarrow a+b+c=-3$ .....$\left(1\right)$

$\Rightarrow ab+bc+ca=0$ ......$\left(2\right)$

$\Rightarrow abc=-(-1)=1$ ......$\left(3\right)$

From $\left(3\right)$ we have

$c=\frac{1}{ab}$ .....$\left(4\right)$

From $\left(1\right)$

$a+b=-3-c$

$\Rightarrow a+b=-3-\frac{1}{ab}$ from $\left(4\right)$

Again from $\left(3\right)$

$\Rightarrow ab+bc+ca=0$

$\Rightarrow ab+c(b+a)=0$

$\Rightarrow ab+\frac{1}{ab}(-3-\frac{1}{ab})=0$

$\Rightarrow {\left(ab\right)}^3-3ab-1=0$

Put $x=ab\Rightarrow$ the equation ${\left(ab\right)}^3-3ab-1=0$

becomes $x^3-3x-1=0$ is the required equation.