Solve x for
I thought of taking to
I am just 9 th grader. I have no idea about solving a cubic equation.
I also think that there should be a way to solve it with the help of logarithmic identities.
Solve x for
I thought of taking to
I am just 9 th grader. I have no idea about solving a cubic equation.
I also think that there should be a way to solve it with the help of logarithmic identities.
I agree up to $\displaystyle \begin{align*} 4 = x^3 - x^2 + 12x + 36 \end{align*}$. Then you need to make use of the null factor law, which states if the product of two terms is 0, then at least one of them is 0.
So you need to set your equation equal to 0 and factorise it. Then you can set each factor equal to 0 and solve for x.
You can also use the rational root theorem: "If the rational number is a root of the polynomial , then n evenly divides the leading coefficient, , and m evenly divides the constant term, ."
Of course, there is no guarantee that a polynomial equation has a rational root but if it does, that limits the possible roots. Here, your equation is [tex]x^3- x^2+ 12x+ 32= 0[tex]. The leading coefficient is 1 so the denominator of any rational root must evenly divide 1- i.e. must be so any rational root must be an integer. The constant term is 32 so if there is a rational root it must be , [tex]\pm 2[tex], , , , or . Try those and see if any work.
For the cubic equation you are solving you should be able to factorize it so that it looks like
When you expand that you will see that -abc is the only term that is not multiplied by x or x^{2} or x^{3}
In your equation the only term not multiplied by x or x^{2} or x^{3} is 32 so -abc=32
If a, b and c are whole numbers (which is not always true but usually is for grade 9 maths) then in order for them to multiply together to give 32 they have to be the factors of 32. The only factors of 32 are: 1, -1, 2, -2, 4, -4, 8, -8, 16, -16, 32, -32 so if you try putting those into the equation you will find some of them which make the equation equal to zero. If you tried 4, for example, and found that putting that into the equation gave an answer of 0 then you know that 4 is a solution of the equation and (x-4) is a factor.
That is the usual way which pre-university students use to solve cubic equations. I have never heard of using logarithms to solve a cubic equation. There is a way to exactly calculate the roots of a cubic equation but it is horribly complicated
For a cubic equation in the form
A cubic with real coefficients may have 1, 2, or 3 distinct real roots. This particular cubic has only 1.
And that root is not approximately 5.2
$5.2^3 - 5.2^2 + 12 * 5.2 + 32 = 140.608 - 27.04 + 62.4 + 32 >> 0.$
In fact, the solution is an irrational number between -1 and -2.
$(-1)^3 - (-1)^2 + 12 * (-1) +32 = -1 - 1 - 12 + 32 = 18.$
$(-2)^3 - (-2)^2 + 12 * (-2) +32 = -8 - 4 - 24 + 32 = - 4.$
And so the answer is not - 4.
This seems like a very strange problem to give to someone who has not studied polynomials. I wonder whether once again you have not given us the original problem.