Q.(a+2)x2 + 2(a+1)x + a=0..The no.of integral valyes of 'a' for which the roots will be integers?
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5 Answers
49{\color{blue} x}{\color{red} =}{\color{green} \frac{-2(a+1)\pm \sqrt{4(a+1)^2-4a(a+2)}}{2(a+2)}}
or,{\color{blue} x}{\color{red} =}{\color{green} \frac{-(a+1)\pm \sqrt{(a+1)^2-a(a+2)}}{a+2}}
or,{\color{blue} x}{\color{red} =}{\color{green}-1+ \frac{1\pm \sqrt{1}}{a+2}}.... [supposing a≠-2....[1]]
so, either, {\color{blue} x}{\color{red} =}{\color{green}-1+ \frac{2}{a+2}}
giving, a=0
or, {\color{blue} x}{\color{red} =}{\color{green}-1+ \frac{0}{a+2}}
giving any value of a except -2
so both roots are integers at a=0.....[1][1][1][1][1]
49vry sry... i was indeed wrong..
the roots are integer at a=0,-1,-3,-4.....so 4 pts..
u kno i missed out my kvpy due to errors like this...jpg)
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49yeeeeeeeeeeeee...............so the q's solved and the thread closed...[1][1][1]