If 9x2+2x−3−4(3x2+2x−2)+27=0,thentheproductofallpossiblevaluesofx$ is
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Official Correct Answer: D. The equation given is:
$$9^{x^2 + 2x - 3} - 4\left(3^{x^2 + 2x - 2}\right) + 27 = 0$$
Hmm, okay. Note that both terms have exponents with base 3 or 9. Since 9 is a power of 3, maybe I can rewrite everything in terms of base 3 to simplify. That seems like a good starting point.
So, I know that 9 is equal to 3 squared, right? So, 9 can be written as 3². Therefore, 9 raised to any power is the same as 3 raised to twice that power. Let me apply that to the first term.
The first term is 9^{x² + 2x - 3}. Let me rewrite that:
$$9^{x^2 + 2x - 3} = (3^2)^{x^2 + 2x - 3} = 3^{2(x^2 + 2x - 3)}$$
Simplifying the exponent:
$$2(x^2 + 2x - 3) = 2x^2 + 4x - 6$$
So, the first term becomes 3^{2x² + 4x - 6}.
Now, the second term is 4*(3^{x² + 2x - 2}). if I can express this in terms of the same exponent as the first term or something related.
Wait, the exponent in the second term is x² + 2x - 2. If I compare that to the exponent in the first term after substitution, which is 2x² + 4x - 6, Note that 2x² + 4x - 6 is exactly twice (x² + 2x - 3). But the exponent in the second term is x² + 2x - 2, which is similar but not exactly half.
If I factor out a 2 from the exponent in the first term, I get 2*(x² + 2x - 3). But the exponent in the second term is x² + 2x - 2, which is x² + 2x - 3 + 1. So, it's like (x² + 2x - 3) + 1. Hmm, not sure if that helps directly.
Maybe I can set a substitution variable to simplify this. Let me let y = 3^{x² + 2x - 3}. That seems like a good substitution because both terms have exponents related to x² + 2x - 3 or similar.
So, let me define:
$$y = 3^{x^2 + 2x - 3}$$
Now, let's express each term in the original equation in terms of y.
First term: 9^{x² + 2x - 3} = 3^{2(x² + 2x - 3)} = (3^{x² + 2x - 3})² = y².
Second term: 4*(3^{x² + 2x - 2}) = 4*(3^{(x² + 2x - 3) + 1}) = 4*(3^{x² + 2x - 3} * 3^1) = 4*3*y = 12y.
Third term: 27, which is 3³, but I don't know if that helps yet.
So, substituting back into the original equation:
$$y² - 12y + 27 = 0$$
Now, that's a quadratic equation in terms of y. Let me write that down:
$$y² - 12y + 27 = 0$$
I can solve this quadratic equation for y. Let's use the quadratic formula. The quadratic is in the form ay² + by + c = 0, where a = 1, b = -12, c = 27.
The quadratic formula is:
$$y = \frac{-b \pm \sqrt{b² - 4ac}}{2a}$$
Plugging in the values:
$$y = \frac{-(-12) \pm \sqrt{(-12)^2 - 4*1*27}}{2*1}$$
Simplify:
$$y = \frac{12 \pm \sqrt{144 - 108}}{2}$$
Calculate the discriminant:
144 - 108 = 36
So,
$$y = \frac{12 \pm \sqrt{36}}{2}$$
$$y = \frac{12 \pm 6}{2}$$
So, two solutions:
1. $$y = \frac{12 + 6}{2} = \frac{18}{2} = 9$$
2. $$y = \frac{12 - 6}{2} = \frac{6}{2} = 3$$
So, y can be either 9 or 3.
But remember, y was defined as:
$$y = 3^{x^2 + 2x - 3}$$
So, we have two equations:
1. $$3^{x^2 + 2x - 3} = 9$$
2. $$3^{x^2 + 2x - 3} = 3$$
Let me solve each equation for x.
Starting with the first equation:
$$3^{x^2 + 2x - 3} = 9$$
Since 9 is 3², we can write:
$$3^{x^2 + 2x - 3} = 3^2$$
Since the bases are equal, the exponents must be equal:
$$x^2 + 2x - 3 = 2$$
Subtract 2 from both sides:
$$x^2 + 2x - 5 = 0$$
Now, solve this quadratic equation for x.
Using the quadratic formula again:
$$x = \frac{-b \pm \sqrt{b² - 4ac}}{2a}$$
Here, a = 1, b = 2, c = -5.
So,
$$x = \frac{-2 \pm \sqrt{(2)^2 - 4*1*(-5)}}{2*1}$$
Simplify:
$$x = \frac{-2 \pm \sqrt{4 + 20}}{2}$$
$$x = \frac{-2 \pm \sqrt{24}}{2}$$
Simplify \sqrt{24} as 2*