How do I solve this?

I'm not good enough to solve this. Sorry!

$$ 3A^2 - 2A^3 = 1.60 $$

  • $\begingroup$ Look up Cardano's formula. $\endgroup$ Commented Aug 25, 2014 at 6:49
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    $\begingroup$ This question appears to be off-topic because it can be answered by googling for a formula and plugging in the numbers. $\endgroup$ Commented Aug 25, 2014 at 6:49
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    $\begingroup$ Check this out !! $\endgroup$
    – creative
    Commented Aug 25, 2014 at 6:52
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    $\begingroup$ If $A$ is a matrix, mention that in your question $\endgroup$
    – Vikram
    Commented Aug 25, 2014 at 6:56
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    $\begingroup$ A being a matrix is not really a question, as the right hand side clearly indicates, we are dealing with a $1\times 1$ matrix. $\endgroup$
    – Barkas
    Commented Aug 25, 2014 at 7:04

3 Answers 3


Instead of just typing the answer, here a short explanation how this is done:

First of all, you should try rearranging things, so you have a $0$ on the right hand side (or the left hand side, as you like): $$ 3A^2-2A^3=1.60 \Leftrightarrow 3A^2-2A^3-1.6 =0 $$ In the next step you should try factorizing the left hand side (i.e. write is as a product) As this is not easily possible here you have the following alternatives:

  • if the highest exponent of your variable is $2$ you can use the following formula
  • if the highest exponent of your variable is $3$ you still can find a solution using Cardano's formula
  • If the highest exponent is higher than $3$ then things turn out to be kind of tedious. In fact, there are still formulas for the case of highest exponent being $4$ but it is proven that factorizing the equation in radicals is not in general possible for degrees of $5$ and higher. In those cases you could "guess" a solution and then do a polynomial long division or use numeric methods, such as the newtown method (among many others)


“Dont give a child a fish but show him how to fish” Mao Tse-tung said or wrote.

Let us take the equation $3A^2 - 2A^3 = b$. If I tell you that, if $b=1.5999$, the real solution is $$\frac{1}{20} \left(10-100 \sqrt[3]{\frac{5}{10999-\sqrt{95978001}}}-\sqrt[3]{\frac{1}{5} \left(10999-\sqrt{95978001}\right)}\right)$$ but that, for $b=1.6001$, the real solution is $$\frac{1}{20} \left(10-100 \sqrt[3]{\frac{5}{11001-\sqrt{96022001}}}-\sqrt[3]{\frac{1}{5} \left(11001-\sqrt{96022001}\right)}\right)$$ what does this bring you ? Just nothing !

Take a look at Cardano method and learn it; this will bring you a lot for all your life !


In addition to using Cardano's formula, polynomial long division, and numerical methods, as other answers suggested, let me suggest a simple method for determining rational roots of a polynomial.

Consider a polynomial $$ p(x)=a_nx^n+...+a_0 $$

The Rational Root Theorem states that if the polynomial $p(x)$ has a rational root, it maust be a fraction $p/q$, where p is a factor of the trailing constant $a_0$ and $q$ is a factor of the leading coefficient $a_n$.

For example, in case of the polynomial $$ p(x)=2x^4 − 11x^3 − 6x^2 + 64^x + 32 $$ the factors of the leading coefficient $a_4=2$ are $\{1,2\}$, whereas the factors of the constant term $a_0=32$ are $\{1,2,4,8,16,32\}$. Therefore, the possible rational roots $x_i$ satisfy the condition $$ x_i\in±\{½, 1, 2, 4, 8, 16, 32\} $$ The Rational Root Theorem is relevant only to rational roots, but it is often helpful in reducing a higher order polynomial equation to a lower order, as it is often the case that a polynomial has a rational root -- at least in the context of undergraduate algebra.

Once you have reduced the order of the polynomial equation by factoring a term in form $(x-x_i)$, where $x_i$ is a root of the original polynomial, you can again attempt to use the methods available at your disposal. In your case, if you find one rational root, you can reduce the equation to a quadratic one and use the quadratic formula for its solution.


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