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tresset_1 [31]
3 years ago
14

6x^2+2x=3 round to the nearest hundreths

Mathematics
1 answer:
vredina [299]3 years ago
4 0

<u>x= 0.56, x= -0.89</u>

<u />

6x^2+2x=3\\Simplify =  6x^2+2x-3=0\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\mathrm{For\:}\quad a=6,\:b=2,\:c=-3:\quad x_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\cdot \:6\left(-3\right)}}{2\cdot \:6}\\x=\frac{-2+\sqrt{2^2-4\cdot \:6\left(-3\right)}}{2\cdot \:6}:\quad \frac{-1+\sqrt{19}}{6}\\x=\frac{-2-\sqrt{2^2-4\cdot \:6\left(-3\right)}}{2\cdot \:6}:\quad -\frac{1+\sqrt{19}}{6}\\

x=\frac{-1+\sqrt{19}}{6},\:x=-\frac{1+\sqrt{19}}{6}

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and then split the fraction,

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Let's factor the 5 out of the intergral,
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Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

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\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
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\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
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