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Solve \(\dfrac{3 x^{2}+2 x+3}{1}+\dfrac{12}{3 x^{2}+2 x+3}=7\)
in Mathematics by Platinum (101k points) | 201 views

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Best answer
let \(3 x^{2}+2 x+3=a\)
So \(\frac{a}{1}+\frac{12}{a}=7\)
multiply by a
&a\left(\frac{a}{1}\right)+\left(1 \frac{2}{2}\right) a=7 a \\
&a^{2}-7 a+12=0 \\
&a^{2}-4 a-3 a+12=0 \\
&a(a-4)-3(a-4)=0 \\

&\text { but } a=3 x^{2}+2 x+3 \\
&\therefore 3 x^{2}+2 x=0 \\
&\Rightarrow x(3 x+2)=0 \\
&\Rightarrow x=0 \text { or } x=-2 / 3 \\
&\text { and } \\
&3 x^{2}+2 x+3=4 \\
&3 x^{2}+2 x-1=0 \\
&3 x^{2}+3 x-x-1=0 \\
&(3 x-1)(x+1)=0 \\
&\Rightarrow x=-1 \text { or } x=1 / 3
by Platinum (101k points)

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