Question

If the equation $x^4-4x^3+a{x}^2+bx+1=0$ has four positive roots then (a,b)is

• A. (-4,6)
• B. (6,-4)
• C. (-4,-6)
• D. (4,-6)

Answer 1

The correct option is B (6,-4)

Let, $\alpha ,\beta ,\gamma ,\delta$ be the roots of the equation
$x^4-4x^3+a{x}^2+bx+1=0$ ...(1)
Sum of the roots $=\alpha +\beta +\gamma +\delta =4$
Product of the roots $=\alpha \beta \gamma \delta =1$
$\frac{\alpha +\beta +\gamma +\delta }{4}={\left(\alpha \beta \gamma \delta \right)}^{1/4}$
Above equation shows that A.M. of the roots is equal to their G.M.
Therefore, all roots are equal.
Therefore, $\alpha =\beta =\gamma =\delta =1$
Equation (1) can be written as:
$x^4-4x^3+a{x}^2+bx+1={(x-1)}^4$
$\Rightarrow x^4-4x^3+a{x}^2+bx+1=x^4-4x^3+6x^2-4x+1$
On Comparing both sides, we get
$a=6,b=-4$
Ans: B