Question

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

• A. $2$
• B. $1$
• C. $10$
• D. $3$

Answer 1

The correct option is C

$10$

Step 1: Use relation between roots and coefficient of equation to get relation

Given: The equation $x^4-4x^3+a{x}^2+bx+1=0$ has four positive roots.

Let the positive roots are $p,q,r$ and $s$.

We know that, Sum of roots$=-\frac{coefficientof{x}^3}{coefficientof{x}^4}$

$\Rightarrow$ $p+q+r+s=4$

$\Rightarrow$ $\frac{p+q+r+s}{4}=1$ ……$\left[1\right]$

We know that, Product of roots$=\frac{cons\tan tterm}{coefficientof{x}^4}$

$\Rightarrow$ $p.q.r.s=1$

$\Rightarrow$ ${\left(p.q.r.s\right)}^{\frac{1}{4}}=1$ …..$\left[2\right]$

Step 2: Compute the value of $a$ and $b$. using concept of A.M. and G.M.

From equation $1$ and $2$

${\left(p.q.r.s\right)}^{\frac{1}{4}}=\frac{p+q+r+s}{4}$

$\Rightarrow$ Geometric mean$=$Arithmetic mean

$\Rightarrow$ $p=q=r=s$

Putting values in equation$1$.

$\Rightarrow$ $p=q=r=s=1$

Therefore, the given equation can be written as:

$\Rightarrow$ ${\left(x-1\right)}^4=0$

$\Rightarrow$ ${\left(x-1\right)}^2{\left(x-1\right)}^2=0$

$\Rightarrow$ $x^4-4x^3+6x^2-4x+1=0$

Comparing with the given equation $x^4-4x^3+a{x}^2+bx+1=0$.

$\Rightarrow$ $a=6$

$\Rightarrow$ $b=-4$

Step 3: Compute the value of $\left|a-b\right|$.

$$\begin{gathered} \left|a-b\right| \\ =\left|6-(-4)\right| \\ =\left|6+4\right| \\ =10 \end{gathered}$$

Hence, the value of $\left|a-b\right|$ is $10$.