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I got this function right here and my teacher wants me to find all real number zeros

$$10x^2+20x+19x+97^1.$$

I looked up this video on how to find it and they were using the $P/Q$ and I found the factors of $97$ and $10$ and them divided them and those were supposed to be the possible zeros, I then did the same thing they did and none of the numbers I got got me zero when I plugged it in, maybe I'm just doing it wrong?

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    $\begingroup$ You equation is oddly written. Why have you written: $10x^2+20x+19x+97^1$ as opposed to $10x^2+39x+97$? $\endgroup$ – Ian Miller Jan 9 '16 at 6:46
  • $\begingroup$ That formula looks fishy. Why $97^1 = 97$ and $20x + 19x = 39x$. Is that just obfuscation or an error putting the question into Math.SE? $\endgroup$ – mvw Jan 9 '16 at 6:47
  • $\begingroup$ It was an error $\endgroup$ – Hooga Jan 9 '16 at 7:58
  • $\begingroup$ You say it was an error but have not changed anything $\endgroup$ – Quality Jan 9 '16 at 19:54
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If your polynomial is indeed what you have written, then

$$10x^{2}+39x+97=0$$

Is simply a quadratic polynomial, so we can apply the quadratic formula, that is

$$x=\frac{ -39 \pm\sqrt{(39)^{2}-4(10)(97)}}{20}$$

Which will yield two complex solutions , no real so most likely you wrote this incorrectly

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  • $\begingroup$ It's supposed to be a function from my birthday so, 10/20/1997 the 10 the 20 the 19 and the 97 $\endgroup$ – Hooga Jan 9 '16 at 7:59
  • $\begingroup$ @hooga but there is no difference between 39x and 20x+19x $\endgroup$ – Quality Jan 9 '16 at 19:54
  • $\begingroup$ Did you want 10x^{2}+20x+1997? $\endgroup$ – Quality Jan 9 '16 at 19:56
  • $\begingroup$ Would that make more sense? $\endgroup$ – Hooga Jan 11 '16 at 1:06
  • $\begingroup$ I don't know it's your question .. $\endgroup$ – Quality Jan 11 '16 at 2:41

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