1.
In each of these questions two equations numbered I & II are given. You have to solve both the equations and give answer.I. $$2a^{2}+3a+1=0$$
II. $$12b^{2}+7b+1=0$$
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By: anil on 05 May 2019 01.32 am
$$2a^{2}+3a+1=0$$ => $$2a^{2}+2a+a+1=0$$ => 2a(a+1) + 1(a+1) = 0 So, a = -1/2 or -1 $$12b^{2}+7b+1=0$$ => $$12b^{2}+3b+4b+1=0$$ => 3b(4b+1)+1(4b+1)=0 So, b = -1/3 or -1/4 Hence a
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