Question

If the zeros of the polynomial $x^3-3x^2+x+1$ are $a-b,a,a+b$, then the value of $a+b$ is

• A. $\sqrt{2}+1$
• B. $1-\sqrt{2}$
• C. $\sqrt{3}-1$
• D. Either $a$ or $b$

Answer 1

Answer: (A), (B)

$\sqrt{2}+1$
$1-\sqrt{2}$

Comparing the given polynomial $x^3-3x^2+x+1$ with $A{x}^3+B{x}^2+Cx+D$. we get
$A=1,B=-3,C=1,D=1$
Here, the zeros of the given polynomial are $a-b,a,a+b$.
$\text{Sum of zeros}=-\frac{B}{A}=-\frac{-3}{1}=3$
$\therefore a-b+a+a+b=3$.
$\Rightarrow 3a=3$.
$\Rightarrow a=1$.

$\text{Product of zeros}=-\frac{D}{A}=-\frac{1}{1}=-1$

$\therefore (a-b)\left(a\right)(a+b)=-1$.
$\Rightarrow a(a^2-b^2)=-1$.
since, $a=1$
$\Rightarrow (1^2-b^2)=-1$.
$\Rightarrow b^2=2$
$\Rightarrow b=\sqrt{2}$ or $-\sqrt{2}$
value of $a+b=\sqrt{2}+1$ or $1-\sqrt{2}$