Q5
If \( 9 ^ { x ^ { 2 } + 2 x - 3 } - 4 \left( 3 ^ { x ^ { 2 } + 2 x - 2 } \right) + 27 = 0 \), then the product of all possible values of \( x \) is
Answer: B
Key idea: Substitute \(t = 3^{x^2+2x-3}\) to get a quadratic.
Then \(9^{x^2+2x-3} = t^2\) and \(3^{x^2+2x-2} = 3t\). The equation becomes
\[t^2 - 12t + 27 = 0 \;\Rightarrow\; (t-3)(t-9) = 0 \;\Rightarrow\; t = 3 \text{ or } 9.\]- \(t = 3\): \(x^2 + 2x - 3 = 1\), i.e. \(x^2 + 2x - 4 = 0\). Discriminant \(4 + 16 > 0\), two real roots, product \(-4\).
- \(t = 9\): \(x^2 + 2x - 3 = 2\), i.e. \(x^2 + 2x - 5 = 0\). Discriminant \(4 + 20 > 0\), two real roots, product \(-5\).
The four roots are distinct, and their product is \((-4)(-5) = 20\).
Hence, the answer is (B) 20.