Instructions

In each of these questions two equations numbered I & II are given. You have to solve both the equations and give answer.

Question 82

I. $$2a^{2}+3a+1=0$$
II. $$12b^{2}+7b+1=0$$

Solution

$$2a^{2}+3a+1=0$$ => $$2a^{2}+2a+a+1=0$$ => 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<b.

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I. $$2a^{2}+3a+1=0$$II. $$12b^{2}+7b+1=0$$