# How do you find the real solutions of the polynomial 9x^3=5x^2+x?

Mar 17, 2018

#### Answer:

$x = 0 , x = \frac{5}{18} \pm \frac{1}{18} \sqrt{61}$

#### Explanation:

$\text{rearrange into standard form}$

$\Rightarrow 9 {x}^{3} - 5 {x}^{2} - x = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\Rightarrow x \left(9 {x}^{2} - 5 x - 1\right) = 0$

$\text{equate each factor to zero and solve for x}$

$\Rightarrow x = 0 \text{ or } 9 {x}^{2} - 5 x - 1 = 0$

$\text{solve "9x^2-5x-1=0" using the "color(blue)"quadratic formula}$

$\text{with "a=9,b=-5" and } c = - 1$

$x = \frac{5 \pm \sqrt{25 + 36}}{18}$

$\textcolor{w h i t e}{x} = \frac{5 \pm \sqrt{61}}{18}$

$\Rightarrow x = \frac{5}{18} \pm \frac{1}{18} \sqrt{61}$

$\Rightarrow x = 0 , x = \frac{5}{18} \pm \frac{1}{18} \sqrt{61} \leftarrow \textcolor{b l u e}{\text{real solutions}}$