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How would you find $x$ in:

$e^{x^2+4x-7}(6x^2+12x+3)=0$

I don't know where to begin. Can you do the following?

$e^{x^2+4x-7}=1/(6x^2+12x+3)$

and then find $ln$ for both sides?

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1 Answer 1

up vote 4 down vote accepted

$e^{x^2+4x-7}(6x^2+12x+3)=0 \Rightarrow e^{x^2+4x-7}=0 \text{ or } \ 6x^2+12x+3=0$

$$\text{It is known that } e^{x^2+4x-7} \text{ is non-zero }$$

therefore,you have to solve :

$$6x^2+12x+3=0$$

The solutions are:

$$x=-1-\frac{1}{\sqrt{2}} \\ x=-1+\frac{1}{\sqrt{2}}$$

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