Precalculus
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# How do you find the real solutions of the polynomial 9x^2+5x^3=6x^4?

Jun 10, 2017

$x = 0 , x \approx - 0.88 , x \approx 1.71$

#### Explanation:

$\text{rearrange the polynomial, equating to zero}$

$\Rightarrow 6 {x}^{4} - 5 {x}^{3} - 9 {x}^{2} = 0$

$\Rightarrow {x}^{2} \left(6 {x}^{2} - 5 x - 9\right) = 0 \leftarrow \text{ common factor}$

$\text{factor " 6x^2-5x-9" using the "color(blue)"quadratic formula}$

$\text{with " a=6,b=-5" and } c = - 9$

$x = \frac{5 \pm \sqrt{25 + 216}}{12} = \frac{5 \pm \sqrt{241}}{12}$

$\Rightarrow x \approx 1.71 \text{ or " x~~-0.88 } \mathmr{and} x = 0$

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