• # question_answer 21) 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 ${{x}^{2}}+ax+b=0$ are [EAMCET 1987] A) $8,\,-1$ B) - 9, 2 C) $-8,-2$ D) 9, 1

8, 2 are the roots of ${{x}^{2}}+ax+\beta =0$ \ $8+2=10=-a$,$8.2=16=\beta$ i.e. $a=-10,\beta =16$ 3, 3 are the roots of ${{x}^{2}}+\alpha x+b=0$ \ $3+3=6=-\alpha ,\,\,3.3=b$ i.e. $\alpha =-6,b=9$ Now, ${{x}^{2}}+ax+b=0$becomes ${{x}^{2}}-10x+9=0$ or$(x-1)(x-9)=0\Rightarrow x=1,\,9$.