Instructions

In each of these question two equations I & II with variables a & b are given You have to solve both the equations to find the values of a & b
Mark answer if
a) a<b

b) $$a\leq b$$

c) relationship between a & b cannot be established

d) a>b

e) $$a\geq b$$

Question 19

I.$$16a^{2}=1$$
II.$$3b^{2}+7b+2=0$$

Solution

$$16a^{2}=1$$
Solving we get, a=-.25, a=+.25

$$3b^{2}+7b+2=0$$
Solving we get, b= -2. b = -1/3

Hence, a>b. Option D is correct.

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I.$$16a^{2}=1$$ II.$$3b^{2}+7b+2=0$$

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I.$$16a^{2}=1$$
II.$$3b^{2}+7b+2=0$$