# Quadratic equation formula help / simplification

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

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.

• 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 0:11
• 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 – gimusi Mar 17 '18 at 0:12
• 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 0:15
• well, $\frac {-10}6$ reduces to $\frac 13(-5)$ and $1060 = 4*265$ so $\sqrt{1060} = \sqrt{4*265} = 2\sqrt{265}$ 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 0:41

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

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

• so is my book wrong ? – italy Mar 17 '18 at 0:12
• Either your book is wrong, or you misread the question or made a typo here. This is the correct answer to this specific equation. – vrugtehagel Mar 17 '18 at 0:14
• For given the quadratic equation this is the solution! Maybe it is symply a typo. – gimusi Mar 17 '18 at 0:14
• okay thank you for that! – italy Mar 17 '18 at 0:16
• @italy You are welcome! Bye – gimusi Mar 17 '18 at 0:17

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

• That's consistant with the OP's answer. The OP just didn't simplify it enough. So that's not "wrong". That's just not finished. – fleablood Mar 17 '18 at 0:43
• @Fleablood: He didn't include $-10$ with the fraction. Now I see it was a typo. I'll edit my answer. Thanks. – Mathew Mahindaratne Mar 17 '18 at 0:49