Question

If $\{8,2\}$ are the roots of $x^2+ax+\beta =0$ and $\{3,3\}$ are the roots of $x^2+\alpha x+b=0$, then the roots of the equation $x^2+ax+b=0$ are:

• A. $1,-1$
• B. $-9,2$
• C. $-8,-2$
• D. $9,1$

Answer 1

Answer: (D)

$9,1$

$8,2$ are the roots of $x^2+ax+\beta =0$
$\Rightarrow a=-(8+2)=-10$

$3,3$ are the roots of $x^2+\alpha x+b=0$
$\Rightarrow b=\left(3\right)\left(3\right)=9$

$x^2+ax+b=0$
$\Rightarrow x^2-10x+9=0$

$\Rightarrow x^2-9x-x+9=0$
$\Rightarrow x(x-9)-1(x-9)=0$
$\Rightarrow (x-9)(x-1)=0$
$\therefore x=9orx=1$