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
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
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.
HINT
We have
$$a^3-6a^2+9a-4=a^3-2a^2+a-4a^2+8a-4=a(a-1)^2-4(a-1)^2$$
$a^3-6a^2+9a-4=0$
$a(a^2+6a+9)=4$
$a(a-3)^2=4\times 1^2$
$a_1=4$
$\frac{a^3-6a^2+9a-4}{a-4}= a^2-2a+1$
$a^2-2a+1=0$
$(a-1)^2=0$
$a_2=a_3=1$
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
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".