anonymous
  • anonymous
Solve the following quadratic for its roots. 3(x+4)=-14/x. I'm having a hard time remember how to do this. Can someone please explain this to me?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
oh thank you
tkhunny
  • tkhunny
Now you still don't know how to do it. What was that @jnn452

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johnweldon1993
  • johnweldon1993
yeah...jnn452 seems to be doing that a lot lately...
whpalmer4
  • whpalmer4
\[3(x+4) = -14/x\]Let's turn this into the standard form \(ax^2+bx+c=0\). Multiply both sides by x, then add 14 to both sides: \[3x(x+4)+14 = 0\]Expand and collect like terms\[3x^2+12x+14=0\]This gives us \(a=3, b=4, c=14\) for the quadratic formula:\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-12\pm\sqrt{(12)^2-4(3)(14)}}{2(3)}\] It should be clear at this point that @jnn452's answer is incorrect. We have two complex roots because the discriminant (\(b^2-4ac\)) is negative, so we'll be getting square roots of negative numbers. \[x = \frac{-12\pm\sqrt{144-168}}{6} = \frac{-12\pm\sqrt{-24}}{6}=\frac{1}{6}(-12\pm 2i\sqrt{6}) \]\[=\frac{1}{3}(-6\pm i\sqrt{6})\] Now we do something that @jnn452 should have done, and check our work :-) \[3(x+4) = 3(\frac{1}{3}(-6+i\sqrt{6})+4) = -6 + i\sqrt{6} +12 =6+i\sqrt{6}\checkmark\] \[\frac{-14}{x} = \frac{-14}{\frac{1}{3}(-6+i\sqrt{6})} = \frac{-42}{-6+i\sqrt{6}} = \frac{-42*(-6-i\sqrt{6})}{(-6+i\sqrt{6})(-6-i\sqrt{6})} \]\[= \frac{-42(-6-i\sqrt{6})}{42} = 6+i\sqrt{6}\checkmark\]The two sides are equal, so the solution checks out.
anonymous
  • anonymous
Thank you so much. I wish I could give you a medal, but after @jnn425 answered my question. I understood it clearly and remember how to do the work. But thank you so much for taking the time. :) Ur so sweet. Thank you

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