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Can you help me solve the quadratic equation $12e^{2x}-32e^x+16$ by factoring please?

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closed as off-topic by user91500, Shailesh, Watson, Claude Leibovici, Nikunj May 27 '16 at 10:22

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  • $\begingroup$ ok i used the methods you gave me and i found x=ln2/3 , x=ln2 $\endgroup$ – super95 May 27 '16 at 8:56
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    $\begingroup$ This is not even an equation unless you add an $=$-sign somewhere... $\endgroup$ – TheAbelian May 27 '16 at 9:12
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$12e^{2x}-32e^x+16=4(3e^{2x}-8e^x+4)=4(3e^x-2)(e^x-2)$

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hint: $4(3e^{2x} - 8e^x+4)=4(3e^x - 2)(e^x-2)$

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Use the substitution $u=e^x$. Your equation will turn into a quadratic equation.

$$12u^2-32u+16$$

You can use the quadratic solution formula to find the roots of this equation.

$$u_{1/2}=\frac{32\pm \sqrt{32^2-4\cdot 12\cdot 16}}{2\cdot 12}$$

You can factor the quadratic equation the following way: $$12(u-u_1)(u-u_2)=12(e^x-u_1)(e^x-u_2)$$

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