Finding roots with seemingly no algebraic way

Question

I have a graph of:

$$y = \frac{x^3 + 2x^2 - 4}{x^2}$$

and I have to find the x-intercept.

So I have the equation $(x^2)(x+2)-4 = 0$

And then I don't know what to do. Not sure if we can use calculators but will use them if there is no other way. Thanks.

Answer 1

$$f(x)=x^3+2x^2-4$$ $$f(0)=-4,f(1)=-1,f(2)=12$$ Since $f(1)f(2)<0$ there is one root in between.

Now this is messy:

$f(1.5)=3.375+4.5-4>0$ So root is in between $1$ and $1.5$

$f(1.25)=1.078>0,f(1.125)=-0.0449<0$

So approximate value of $x\approx1.1(8)$ satisfies your equation, with exact value at $x=1.13039543476\cdots$ seems close, or you may iterate further if you want

Answer 2

Replace $x=\frac{2}{y}$ in equation $x^3+2x^2-4=0$ and get $$ \left(\frac{2}{y}\right)^3+2\left(\frac{2}{y}\right)^2-4=0 $$ and after simplifications $$ y^3-2y-2=0 $$ Now use the Cardamo's formula for cubic equation $y^3+py+q=0$ $$ y=\sqrt[3\,]{-\frac{p}{2}+\sqrt{\left(-\frac{p}{2}\right)^2+\left(-\frac{q}{3}\right)^3}} + \sqrt[3\,]{-\frac{p}{2}-\sqrt{\left(-\frac{p}{2}\right)^2+\left(-\frac{q}{3}\right)^3}} $$ Now (as suggested by André Nicolas) follow the instructions on wikipedia to manipulate the formula and get a algebric representation of roots.