Question

$I.12p^2-7q=-1II.6q^2-7q+2=0$

• A. p < q
• B. p > q
• C. p $\le$ q
• D. p $\ge$ q
• E. p = q

Answer 1

The correct option is A p < q

$\left(I\right)12p^2-7p=-1\Rightarrow 12p^2-7p+1=0\Rightarrow 12p^2-4p-3p+1=0\Rightarrow 4p(3p-1)-1(3p-1)=0\Rightarrow (3p-1)(4p-1)=0\Rightarrow p=\frac{1}{4}or\frac{1}{3}\left(II\right)6q^2-7q+2=0\Rightarrow 6q^2-4q-3q+2=0\Rightarrow 2q(3q-2)-1(3q-2)=0\Rightarrow (3q-2)(2q-1)=0\Rightarrow q=\frac{2}{3}or\frac{1}{2}$
Obviously, $p<q$