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: Two layers of substitution. First let \(u=x^2+2x\) so the equation is in one exponent; then let \(y=3^u\) to turn it into a plain quadratic. Solve, back-substitute, and read off the product via Vieta — no need to actually solve for \(x\).

Step 1: Put \(u = x^2+2x\). The equation \(9^{x^2+2x-3} - 4\cdot 3^{x^2+2x-2} + 27 = 0\) becomes \[ 9^{\,u-3} - 4\cdot 3^{\,u-2} + 27 = 0. \]

Step 2: Since \(9=3^2\), write \(9^{u-3}=3^{2u-6}\). Let \(y = 3^{u}\). Then \(3^{2u-6} = \dfrac{y^2}{3^6} = \dfrac{y^2}{729}\) and \(3^{u-2} = \dfrac{y}{9}\), so \[ \frac{y^2}{729} - \frac{4y}{9} + 27 = 0. \] Multiply by \(729\): \[ y^2 - 324y + 19683 = 0. \]

Step 3: Discriminant \(= 324^2 - 4(19683) = 104976 - 78732 = 26244 = 162^2\), so \[ y = \frac{324 \pm 162}{2} = 243 \ \text{or}\ 81. \]

Step 4: Back-substitute \(y=3^u\): \[ 3^u = 243 = 3^5 \Rightarrow u = 5, \qquad 3^u = 81 = 3^4 \Rightarrow u = 4. \] With \(u = x^2+2x\) this gives two quadratics: \[ x^2+2x-5 = 0, \qquad x^2+2x-4 = 0. \] Each has real roots (discriminants \(24\) and \(20\), both positive), so all four \(x\)-values are valid.

Step 5: By Vieta, product of roots \(=\dfrac{c}{a}\): for the first \(=-5\), for the second \(=-4\). Product of all four roots: \[ (-5)\times(-4) = 20. \] Final answer: \(20\).

Answer 2

Step 1: Introduce a substitution. Let \[ u = x^2 + 2x. \] Then the equation becomes \[ 9^{u-3} - 4\cdot 3^{u-2} + 27 = 0. \]
Step 2: Express everything with base $3$. Since $9 = 3^2$, \[ 9^{u-3} = (3^2)^{u-3} = 3^{2(u-3)} = 3^{2u-6}. \] So we have \[ 3^{2u-6} - 4\cdot 3^{u-2} + 27 = 0. \] Rewrite using negative exponents (or dividing by powers of $3$): \[ 3^{2u-6} = \frac{3^{2u}}{3^6}, \qquad 3^{u-2} = \frac{3^u}{3^2}. \] Thus, \[ \frac{3^{2u}}{3^6} - 4\cdot \frac{3^u}{3^2} + 27 = 0 \;\Rightarrow\; \frac{(3^u)^2}{729} - \frac{4\cdot 3^u}{9} + 27 = 0. \]
Step 3: Quadratic in $3^u$. Let \[ y = 3^u. \] Then \[ \frac{y^2}{729} - \frac{4y}{9} + 27 = 0. \] Multiply through by $729$: \[ y^2 - 4\cdot 81\,y + 27\cdot 729 = 0 \;\Rightarrow\; y^2 - 324y + 19683 = 0. \] Solve: \[ y = \frac{324 \pm \sqrt{324^2 - 4\cdot 19683}}{2}. \] Compute the discriminant: \[ 324^2 = 104976,\quad 4\cdot 19683 = 78732, \] \[ 104976 - 78732 = 26244 = 162^2. \] So \[ y = \frac{324 \pm 162}{2}. \] Hence \[ y_1 = \frac{324 + 162}{2} = \frac{486}{2} = 243,\qquad y_2 = \frac{324 - 162}{2} = \frac{162}{2} = 81. \]
Step 4: Back-substitute for $u$. Recall $y = 3^u$. \[ 3^u = 243 = 3^5 \Rightarrow u = 5, \qquad 3^u = 81 = 3^4 \Rightarrow u = 4. \] But $u = x^2 + 2x$. So we get two quadratics: \[ x^2 + 2x = 5 \Rightarrow x^2 + 2x - 5 = 0, \] \[ x^2 + 2x = 4 \Rightarrow x^2 + 2x - 4 = 0. \]
Step 5: Product of all possible $x$. For a quadratic $ax^2 + bx + c = 0$, product of roots $= \dfrac{c}{a}$. For $x^2 + 2x - 5 = 0$, product of roots \[ P_1 = -5. \] For $x^2 + 2x - 4 = 0$, product of roots \[ P_2 = -4. \] All possible $x$ are the four roots from these two equations, so the product of all possible $x$ is \[ P_1 \cdot P_2 = (-5)\cdot(-4) = 20. \] Therefore, the product of all possible values of $x$ is \(20\).