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I have this quadratic equation,

$ x^{2} + \frac{10}{3}x -\frac{80}{3} = 0 $

I use the quadratic formula to solve and simplify

$-10 \pm \frac{\sqrt{100-(4)(3)(-80)}}{6}$ = $ \frac{-10 \pm \sqrt{1060}}{6}$

my book says it should simplify to

$ \frac{1}{3} ( -5 \pm \sqrt{73}) $

but i cant get this simplification can anyone show me if they can? Thank you.

italy
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  • I'd suggest double checking that you copied the original problem correctly. That $\frac{1}{3} ( -5 \pm \sqrt{73})$ is not the correct answer for the quadratic you wrote down. – sharding4 Mar 17 '18 at 00:11
  • it seems ok, there is only a typo here for the expression $-10 \pm \frac{\sqrt{100-(4)(3)(-80)}}{6}$ but the result is correct – user Mar 17 '18 at 00:12
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    The book solutions are roots of the (different) quadratic $,x^2 + \dfrac{10}{3} x - \dfrac{\color{red}{16}}{3} = x^2 + \dfrac{10}{3} x - \dfrac{80}{\color{red}{15}},$. – dxiv Mar 17 '18 at 00:15
  • well, $\frac {-10}6$ reduces to $\frac 13(-5)$ and $1060 = 4265$ so $\sqrt{1060} = \sqrt{4265} = 2\sqrt{265}$ so $\frac {\pm \sqrt{1060}}6$ reduces to $\frac 13(\pm \sqrt{265})$. That you got 265 and the book got 73 is probably an arithmetic error. (apparently on you book's end) – fleablood Mar 17 '18 at 00:41

2 Answers2

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Your solution seems correct, indeed

$$x^{2} + \frac{10}{3}x -\frac{80}{3} = 0\iff3x^2+10x-80=0$$

$$ \frac{-10 \pm\sqrt{100-(4)(3)(-80)}}{6}=-\frac53\pm\frac{\sqrt{265}}{3}$$

user
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$$x=\frac{-10 \pm \sqrt{100-(4)(3)(-80)}}{6} =\frac{-2\times 5 \pm \sqrt{4\times25-(4)(3)(-80)}}{6} =\frac{1}{6}\left(-2\times 5 \pm 2\sqrt{25+240} \right) =\frac{2}{6}\left(-5 \pm \sqrt{265} \right)=\frac{1}{3}\left(-5 \pm \sqrt{265} \right)$$

Mathew Mahindaratne
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