(b) Given that $3^{2x^{2} - 7x + 7} = 3^{2} \Rightarrow 2x^{2} - 7x + 7 = 2$ $\Rightarrow 2x^{2} + 7x + 5 = 0$ Now, $D = b^{2} - 4ac$ $= ( - 7)^{2} - 4 \times 2 \times 5$ $= 49 - 40 = 9 > 0$ Hence, it has two real roots.
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