How to solve cubic equations to 2 decimal places

Key concept: When an equation won’t factor nicely, numerical solving is the practical route. You hunt for sign changes, then narrow the interval until the decimals settle down.

Step by step

We want the solutions correct to 2 decimal places.

a) $x^3-3x-1=0$

Let

$f(x)=x^3-3x-1.$

Check values:

\quad f(2)=8-6-1=1.$$ So there is a root between 1 and 2. Try closer values: $$f(1.88)\approx 1.88^3-3(1.88)-1\approx -0.00,$$ so one solution is $$\boxed{x\approx 1.88}.$$ Because this cubic has three real roots, the others are approximately $$\boxed{x\approx -1.53 \text{ and } x\approx -0.35}.$$ ## b) $x^3+4=3x^2+x$ Rearrange: $$x^3-3x^2-x+4=0.$$ Now test simple values: $$f(2)=8-12-2+4=-2, \quad f(3)=27-27-3+4=1.$$ So one root lies between 2 and 3. Checking more closely gives $$\boxed{x\approx 2.72}.$$ Also, the cubic factors as $$x^3-3x^2-x+4=(x-2)(x^2-x-2)=(x-2)(x-2)(x+1),$$ so the exact solutions are $$x=2 \text{ (double root) }, \quad x=-1.$$ To 2 decimal places: $$\boxed{x=2.00 \text{ and } x=-1.00}.$$ ### Final answers - a) $x\approx -1.53,\,-0.35,\,1.88$ - b) $x=-1.00,\,2.00$ ### Pitfall alert For cubic equations, don’t stop after finding one root unless the question clearly asks for just one. Also, if the equation factors, it is worth checking because exact roots are cleaner than decimal approximations. Be careful with repeated roots like $x=2$ here: the value appears twice in the factorisation. ### Try different conditions If the coefficients changed slightly, the strategy would be the same: rearrange to one side, test values to locate roots, then refine with a calculator or factorisation if possible. For example, changing $x^3-3x-1=0$ to $x^3-3x-2=0$ gives a different set of roots, but the sign-check method still works. ### Further reading cubic equation, numerical solution, factorisation