Question

If $9^{x^2+2x-3} - 4\bigl(3^{x^2+2x-2}\bigr) + 27 = 0$, then the product of all possible values of $x$ is:

• A. \(2\)
• B. \(4\)
• C. \(10\)
• D. \(20\)

Hint:

When exponents share a common expression (like $x^2+2x$ here), substitute it with a single variable to simplify. Then, look for ways to convert to a common base and reduce the equation to a quadratic in a new variable such as $3^u$.

Answer 1

The Correct Option is D

Approach: Spot that the equation is quadratic in $3^{x^2+2x}$ if you factor out the powers cleanly, but here is a slicker route: divide the whole equation by $27$ to normalise it, which makes the hidden quadratic in $t=3^{x^2+2x}$ pop out with small coefficients.

Step 1: Let $t = 3^{\,x^2+2x}$. Then, using $9^{x^2+2x}=t^2$ and pulling the constant exponents out as numerical factors: \[ 9^{x^2+2x-3} = \frac{t^2}{9^3} = \frac{t^2}{729},\qquad 3^{x^2+2x-2} = \frac{t}{9}. \] The equation becomes \[ \frac{t^2}{729} - \frac{4t}{9} + 27 = 0. \]

Step 2: Multiply through by $729$ and divide by the common structure: \[ t^2 - 324\,t + 19683 = 0. \] Factor by looking for two numbers multiplying to $19683 = 3^9$ and adding to $324$: those are $243=3^5$ and $81=3^4$. So \[ (t - 243)(t - 81) = 0 \;\Rightarrow\; t = 243 \ \text{or}\ 81. \]

Step 3: Recall $t = 3^{x^2+2x}$: \[ 3^{x^2+2x} = 3^5 \Rightarrow x^2+2x = 5, \qquad 3^{x^2+2x} = 3^4 \Rightarrow x^2+2x = 4. \]

Step 4: So $x^2+2x-5=0$ (roots multiply to $-5$) and $x^2+2x-4=0$ (roots multiply to $-4$). Both quadratics have positive discriminant, so all roots are real and admissible. Product of all possible $x$: \[ (-5)(-4) = 20. \] Final answer: $20$.