Question

If the zero of the polynomial $x^3-3x^2+x+1$ are $a-b,a,a+b$, find '$a$' and '$b$'

• A. $a=1,b=\pm \sqrt{3}$
• B. $a=1,b=\pm \sqrt{2}$
• C. $a=\pm \sqrt{2},b=1$
• D. $a=\pm 1,b=\sqrt{2}$

Answer 1

The correct option is B $a=1,b=\pm \sqrt{2}$

As coefficient of $x^3$ is $1$, if there roots are $\alpha ,\beta ,\gamma$, we have

$x^3-3x^2+x+1=(x-\alpha )(x-\beta )(x-\gamma )$

$=x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +(\alpha )x+\alpha \beta \gamma )$

Now, compare coefficient of similar powers on each side.

First compare sum of roots from coefficient of $x^2$,

$\therefore a-b+a+a+b=i.e3a=\pm 3$ i.e. $a=+1$

zero coefficient of $x$ gives,

$a(a-b)+a(a+b)+(a+b)(a-b)=1$

i.e $a(a-b+a+b)+a^2-b^2=1$ or $3a^2-b^2=1$

and as $a=1$

$b^2=3\times (1{)}^2-1$

$b^2=2$

$b=\pm \sqrt{2}$