I have this math problem that I have to solve (Edexcel GCSE 9-1 Mathematics Higher Student Book) which says:
Show that the equation $$x^3+4x=6$$ has a solution between 1.1 and 1.2.
The lesson is on expanding and factorizing, and all these problems are quadratic except this one. I have asked my teacher for a hint and all she says is "look at the problem from a different angle".
How can I prove it (obviously not using the calculator equation solving mode)?
It is enough to apply the intermediate value theorem to the continuous function $p(x)=x(x^2+4)$.
We have $p\left(\frac{11}{10}\right) = \frac{11}{10}\cdot \frac{521}{100}<6 $ and $p\left(\frac{6}{5}\right)=\frac{6}{5}\cdot\frac{136}{25}>6$, hence $p(x)$ takes for sure the value $6$ over the interval $I=(1.1,1.2)$. Since $p(x)$ is a convex function over such interval (due to $p''(x)>0$) the sequence of approximations provided by $$ x_0=\frac{6}{5},\qquad x_{n+1}=2\cdot\frac{3+x^3}{4+3x^2} $$ is quadratically and monotonically convergent to an actual zero of $p(x)-6$ over $I$, by the properties of Newton's method. If we consider $x_1=\frac{591}{520}$, we already have $$ p(x_1)-6<\frac{1}{70}.$$ We may also find an explicit solution. Since $p(x)$ is an increasing function (due to $p'(x)>0$),
the only real zero of $p(x)-6$ is given by
$$ \xi = \alpha+\frac{4}{3\alpha},\qquad \alpha=\sqrt[3]{3+\tfrac{1}{9}\sqrt{921}} $$ by the cubic formula, where $\xi\approx 1.1347284533618458$.