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

Question

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

1 Answer

Impact of this question

2441 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-17T11:15:50

$x = 0, x = \frac{5}{18} \pm \frac{1}{18} \sqrt{61}$ $x=0,x=5/18+-1/18sqrt61$

Explanation:

$rearrange into standard form$ $"rearrange into standard form"$

$\Rightarrow 9 x^{3} - 5 x^{2} - x = 0 \leftarrow in standard form$ $rArr9x^3-5x^2-x=0larrcolor(blue)"in standard form"$

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

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

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

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

$with a = 9, b = - 5 and c = - 1$ $"with "a=9,b=-5" and "c=-1$

$x = \frac{5 \pm \sqrt{25 + 36}}{18}$ $x=(5+-sqrt(25+36))/18$

$x = \frac{5 \pm \sqrt{61}}{18}$ $color(white)(x)=(5+-sqrt61)/18$

$\Rightarrow x = \frac{5}{18} \pm \frac{1}{18} \sqrt{61}$ $rArrx=5/18+-1/18sqrt61$

$\Rightarrow x = 0, x = \frac{5}{18} \pm \frac{1}{18} \sqrt{61} \leftarrow real solutions$ $rArrx=0,x=5/18+-1/18sqrt61larrcolor(blue)"real solutions"$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the real solutions of the polynomial 9x3=5x2+x9x^3=5x^2+x?

1 Answer

Explanation:

Related questions

Impact of this question

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