The question is: how many solutions does this equation have?
$$\frac {2x^3+1.6x}{x^2-1} = 7$$
I don't even have a clue how to approach this…
[Math] How many solutions does this equation have
algebra-precalculus
algebra-precalculus
The question is: how many solutions does this equation have?
$$\frac {2x^3+1.6x}{x^2-1} = 7$$
I don't even have a clue how to approach this…
Best Answer
$$\frac{2x^3+1.6x}{x^2-1}=7$$ $$\frac{2x^3+1.6x}{x^2-1}(x^2-1)=7(x^2-1)$$ $$2x^3+1.6x=7x^2-7$$ $$2x^3-7x^2+1.6x+7=0$$ You are now left with a polynomial equation of degree three - a cubic, in the form of $ax^3+bx^2+cd+d=0$. All you have to do is find the roots. This isn't a special case, unfortunately, so we have to do it the long way. A root $x_k$ is $$x_k=-\frac{1}{3a}\left(b+u_kC+\frac{\Delta_0}{u_kC}\right)$$ where $$u_1=1, u_2=\frac{-1+i\sqrt{3}}{2}, u_3=\frac{-1-i\sqrt{3}}{2}$$ $$C=\sqrt[3]{\frac{\Delta_1\sqrt{\Delta_1^2-4\Delta_0^3}}{2}}$$ Additionally, $$\Delta_0=b^2-3ac, \Delta_1=2b^3-9abc+27a^2d$$ I leave you to calculate $a$, $b$, $c$, and $d$, and the associated quantities. The algebra should be simply enough from there.