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If $$9^{x^{2}+2x-3}-4(3^{x^{2}+2x-2})+27=0$$ then the product of all possible values of x is
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If $$9^{x^{2}+2x-3}-4(3^{x^{2}+2x-2})+27=0$$ then the product of all possible values of x is
Let's assume that $$x^{2}+2x-3 = t$$
$$9^{x^{2}+2x-3}-4(3^{x^{2}+2x-2})+27=0$$ can be written as $$9^t-4(3^{t+1})+27=0$$
$$3^{2t}-12(3^t)+27=0$$
Let's assume that $$3^t = y$$
$$y^2-12y+27=0$$
$$(y-9)(y-3)=0$$
y = 3 or 9 $$\Rightarrow$$ t = 1 or 2
Let's solve when t = 1
$$x^{2}+2x-3 = 1 \Rightarrow x^{2}+2x-4 = 0$$
$$b^2-4ac\ =\ 4+16\ =\ 20$$
Positive, so the equation has real roots.
Product of possible value of x = -4
Let's solve for t = 2
$$x^{2}+2x-3 = 2 \Rightarrow x^{2}+2x-5 = 0$$
$$b^2-4ac\ =\ 4+20\ =\ 24$$
Positive, so the equation has real roots.
Product of possible value of x = -5
The product of all values = 20
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