2x^3 + x^2 - 16 = 0  and how to solve these equations, given that I'm in grade 11? ax^3 + bx^2 + cx + d = 0  , "a" not equal 0

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should know that there is no such a formula to help you to find the roots of an equation whose degree is larger than 2.

There are other methods that help you to find the real roots of an equation. One method specifies to form a set whose elements are irreducible fractions. The numerator of these fractions is one of divisors of constant term and the denominator is one of divisors of leading coefficient such that:

`D_16 {+-1;+-2;+-4;+-8;+-16} `

`D_2 {+-1;+-2}`

The set is formed from the following fractions such that:

`(D_16)/(D_2) = {+-1/2}`

Notice that the only irreducible fraction `x = 1/2`  does not cancel the equation, hence, `x = 1/2`  is not a solution to the given equation.

The graphical method helps you to figure out how many real roots the given equation has such that:

Notice that the graph intersects x axis only one time, hence, the equation has one real solution, `x in (1.5 , 2)`  and two complex solutions.