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

The correct option is 20.

We are given the exponential equation:

$9^{x^2+2x-3}-4\left(3^{x^2+2x-2}\right)+27=0$

Step 1: Simplify the exponential terms
Let $y=x^2+2x$.
Substitute y into the exponent terms:

$9^{y-3}-4\left(3^{y-2}\right)+27=0$

Rewrite base 9 as $3^2$:

${\left(3^2\right)}^{y-3}-4\left(3^{y-2}\right)+27=0$

$3^{2y-6}-4\left(3^{y-2}\right)+27=0$

Multiply the entire equation by $3^6$ to clear denominators, or express in terms of $t=3^y$:

$\frac{3^{2y}}{3^6}-4\cdot \frac{3^y}{3^2}+27=0$

Substitute $t=3^y$ (where t > 0):

$\frac{t^2}{729}-\frac{4t}{9}+27=0$

Multiply the entire equation by 729:

$t^2-4\cdot 81t+27\cdot 729=0$

$t^2-324t+19683=0$

Step 2: Solve the quadratic equation for t
We look for factors of 19683 that sum to 324:
Notice that $81+243=324$ and $81\times 243=3^4\times 3^5=3^9=19683$.

So the equation factors as:

$(t-81)(t-243)=0$

Thus, the roots for t are:

$t=81=3^4$ or $t=243=3^5$

Step 3: Solve for y
Since $t=3^y$:

1. If $3^y=3^4$, then $y=4$
2. If $3^y=3^5$, then $y=5$

Step 4: Solve for x
Recall that $y=x^2+2x$.

Case 1: $x^2+2x=4$

$x^2+2x-4=0$

Let the roots of this equation be $x_1$ and $x_2$. By Vieta's formulas, the product of roots is:

$x_1\cdot x_2=-4$

Case 2: $x^2+2x=5$

$x^2+2x-5=0$

Let the roots of this equation be $x_3$ and $x_4$. By Vieta's formulas, the product of roots is:

$x_3\cdot x_4=-5$

Step 5: Find the product of all possible values of x
The product of all four real solutions x₁, x₂, x₃, and x₄ is:

$\text{Product}=(x_1\cdot x_2)\cdot (x_3\cdot x_4)=(-4)\cdot (-5)=20$

Thus, the product of all possible values of x is 20.