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How to solve the roots of following cubic equation

$$a^3-6a^2+9a-4=0$$

I am solving roots of characteristics equation

so I use casio 991ms calculator so it gave me roots $-0.355301,3.177650,3.1776506$

please help me with this

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    $\begingroup$ Your Casio, or something you entered into it, is wrong. The coefficients add up to $0$ which should be a dead giveaway for one root. In a correct solution you should find all roots are whole numbers! $\endgroup$
    – Oscar Lanzi
    Sep 18, 2018 at 15:50
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    $\begingroup$ Rectify the Casio! $\endgroup$
    – user
    Sep 18, 2018 at 15:50
  • $\begingroup$ Better to do by hand in the old fashioned way. HINT Sum of coefficients = zero means $(x-1)$ is a root.. which you missed two times. $\endgroup$
    – Narasimham
    Sep 18, 2018 at 19:45

5 Answers 5

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you did something wrong, not sure calculator issue or a typo.

For cubic equations, if it's a setup question better to try a simple root first by inspection.

Hint: try $a=1$.

Once you have one value, you can reduce to second order and find the other two roots.

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  • $\begingroup$ it only gave me 2 roots 1,4 how to find 3rd root $\endgroup$
    – Mandarr Sant
    Sep 18, 2018 at 15:56
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    $\begingroup$ Divide out the $a=1$ root by dividing by $a-1$. Solve the quadratic equation by factoring, completing the square, or the quadratic formula for the remaining roots. You should find your "missing root" is actually a double root. The problem is testing your ability to incorporate a double root into your solution, which your book should have told you how to do. $\endgroup$
    – Oscar Lanzi
    Sep 18, 2018 at 15:59
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HINT

We have

$$a^3-6a^2+9a-4=a^3-2a^2+a-4a^2+8a-4=a(a-1)^2-4(a-1)^2$$

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    $\begingroup$ By guessing, since $a=1$ is a root, dividing $a^3-6a^2+9a-4|a-1$ we obtain $a^2-5a+4=(a-1)(a-4)$. $\endgroup$
    – user
    Sep 18, 2018 at 15:56
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    $\begingroup$ how do u entered this equation(while answering) bcoz i used html(while typing question) is there any alternative to(easy way )to enter the equation(while asking question on this site) i am new to this site $\endgroup$
    – Mandarr Sant
    Sep 18, 2018 at 16:00
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    $\begingroup$ @MandarSant Just type "a^3-6a^2+9a-4=a^3-2a^2+a-4a^2+8a-4=a(a-1)^2-4(a-1)^2" and use the marker "$" at the biginning and at the end. $\endgroup$
    – user
    Sep 18, 2018 at 16:02
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    $\begingroup$ @MandarSant If you push on "edit" commend you can see all the full text. As an alternative you can copy that through the right button of the mouse. $\endgroup$
    – user
    Sep 18, 2018 at 16:03
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$a^3-6a^2+9a-4=0$

$a(a^2+6a+9)=4$

$a(a-3)^2=4\times 1^2$

$a_1=4$

$\frac{a^3-6a^2+9a-4}{a-4}= a^2-2a+1$

$a^2-2a+1=0$

$(a-1)^2=0$

$a_2=a_3=1$

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  • $\begingroup$ $a^3-7a^2+11a-5=0$ can you help me with this? $\endgroup$
    – Mandarr Sant
    Sep 20, 2018 at 17:14
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    $\begingroup$ Like previous equation the sum of coefficients of terms of polynomial is 0, i,e. $1-7+11-5=0$, that is $a=1$is a solution.Divide $a^3-7a^2+11a-5$ by (a-1) you get an equation of degree 2 which can be solved easily. $\endgroup$
    – sirous
    Sep 20, 2018 at 19:20
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Get the critical values of the equation by testing for a number that produces the result 0. Then use that equation to divide the polynomial, you'll be left with a quadratic equation. I think you can continue from there

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    $\begingroup$ "Critical values" normally refers to places where the derivative vanishes or doesn't exist. $\endgroup$
    – Cheerful Parsnip
    Sep 18, 2018 at 16:09
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The numbers $1,4$ are roots of your equation. Can you find the rest? In case you want to have a view of the graph you can use a nice software like "desmos".

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  • $\begingroup$ i want 3 roots not only 1 and 4 $\endgroup$
    – Mandarr Sant
    Sep 18, 2018 at 15:55
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    $\begingroup$ There are no more, actually...the $1$ is double. See gimusi factorization, please. $\endgroup$
    – dmtri
    Sep 18, 2018 at 15:59
  • $\begingroup$ how to identified which root is double using calculator $\endgroup$
    – Mandarr Sant
    Sep 18, 2018 at 16:01
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    $\begingroup$ You only think you want three roots. See my comment under another answer. $\endgroup$
    – Oscar Lanzi
    Sep 18, 2018 at 16:02
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    $\begingroup$ Nice question, but I do not know about Casio calculators... $\endgroup$
    – dmtri
    Sep 18, 2018 at 16:03

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