# Approximate Solutions of Equations Question?

## Show that the equation ${x}^{3} - 12 x + 10 = 0$ has three real roots and determine two consecutive numbers between which each of the roots lie. Hence, by applying the half-interval method, approximate the root nearest to the origin to 1dp. Thanks!

Oct 7, 2017

The root near $0$ is approximately $0.9$

#### Explanation:

Let:

$f \left(x\right) = {x}^{3} - 12 x + 10$

Then:

$f \left(- 4\right) = - 64 + 48 + 10 = - 6 < 0$

$f \left(- 3\right) = - 27 + 36 + 10 = 19 > 0$

$f \left(0\right) = 0 - 0 + 10 = 10 > 0$

$f \left(1\right) = 1 - 12 + 10 = - 1 < 0$

$f \left(2\right) = 8 - 24 + 10 = - 6 < 0$

$f \left(3\right) = 27 - 36 + 10 = 1 > 0$

So $f \left(x\right)$ has real zeros in $\left(- 4 , - 3\right)$, $\left(0 , 1\right)$ and $\left(2 , 3\right)$

The root nearest the origin is the one in $\left(0 , 1\right)$

Personally, I would linearly interpolate to find approximate root $\frac{10}{11} \approx 0.9$, but let's bisect the interval to find the approximation...

$f \left(\frac{1}{2}\right) = {\left(\frac{1}{2}\right)}^{3} - 12 \left(\frac{1}{2}\right) + 10 = \frac{1}{8} - 6 + 10 = \frac{33}{8} > 0$

$f \left(\frac{3}{4}\right) = {\left(\frac{3}{4}\right)}^{3} - 12 \left(\frac{3}{4}\right) + 10 = \frac{27}{64} - 9 + 10 = \frac{91}{64} > 0$

$f \left(\frac{7}{8}\right) = {\left(\frac{7}{8}\right)}^{3} - 12 \left(\frac{7}{8}\right) + 10 = \frac{343}{512} - \frac{21}{2} + 10 = \frac{87}{512} > 0$

$f \left(\frac{15}{16}\right) = {\left(\frac{15}{16}\right)}^{3} - 12 \left(\frac{15}{16}\right) + 10 = \frac{3375}{4096} - \frac{45}{4} + 10 = - \frac{1745}{4096} < 0$

So the zero lies between $\frac{7}{8} = 0.875$ and $\frac{15}{16} = 0.9375$, so to one decimal place is $0.9$.

Bonus

A more efficient way of finding the zeros is Newton's method.

Given:

$f \left(x\right) = {x}^{3} - 12 x + 10$

the derivative of $f \left(x\right)$ is:

$f ' \left(x\right) = 3 {x}^{2} - 12$

Then given an approximate zero $x = a$, a better approximation is:

$a - \frac{f \left(a\right)}{f ' \left(a\right)} = a - \frac{{x}^{3} - 12 x + 10}{3 {x}^{2} - 12}$

In our example, we could begin by with $a = 1$.

Then a better approximation is:

$1 - \frac{f \left(1\right)}{f ' \left(1\right)} = 1 - \frac{1 - 12 + 10}{3 - 12} = 1 - \frac{- 1}{- 9} = \frac{8}{9} = 0. \overline{8} \approx 0.9$

We can repeat to get a better approximation:

$\frac{8}{9} - \frac{f \left(\frac{8}{9}\right)}{f ' \left(\frac{8}{9}\right)} = \frac{8}{9} - \frac{\frac{512}{729} - \frac{32}{3} + 10}{\frac{64}{27} - 12}$

$\textcolor{w h i t e}{\frac{8}{9} - \frac{f \left(\frac{8}{9}\right)}{f ' \left(\frac{8}{9}\right)}} = \frac{8}{9} - \frac{\frac{26}{729}}{- \frac{260}{27}}$

$\textcolor{w h i t e}{\frac{8}{9} - \frac{f \left(\frac{8}{9}\right)}{f ' \left(\frac{8}{9}\right)}} = \frac{8}{9} + \frac{1}{270}$

$\textcolor{w h i t e}{\frac{8}{9} - \frac{f \left(\frac{8}{9}\right)}{f ' \left(\frac{8}{9}\right)}} = \frac{241}{270}$

$\textcolor{w h i t e}{\frac{8}{9} - \frac{f \left(\frac{8}{9}\right)}{f ' \left(\frac{8}{9}\right)}} = 0.8 \overline{925} \approx 0.8926$