There is an equation X^3 + X^2 -2x = 0 I have calculated it like below
X^2(X-1) + 2X(X-1) = 0
X(X-1) (X+2) = 0
but how to calculate the values of X here?
There is an equation X^3 + X^2 -2x = 0 I have calculated it like below
X^2(X-1) + 2X(X-1) = 0
X(X-1) (X+2) = 0
but how to calculate the values of X here?
You have a product equal to 0. What can we conclude from this? Why did you convert the left-hand side to a product in the first place?
I don't understand what you say. Could you simplify it, please?
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
There is a basic property of numbers known as the zero product property.
If a product equals zero, then AT LEAST one of the factors must equal zero.
So if [tex]x(x - 1)(x + 2) = 0 \implies WHAT?[/tex]
A common way to find the zeroes of a polynomial is to factor it. (The Fundamental Theorem of Algebra says that every polynomial of degree n > 2 and with real coefficients can be factored into polynomials of lower degree with real coefficients. Unfortunately the theorem gives no guidance on how to do that factoring.)
Do you mean to say, if x(x - 1)(x + 2) = 0 then X = 0, X-1 = 0 and X+2 = 0? Could you clarify it please?
[tex]x = 0 \implies x = 0.[/tex]
[tex]x - 1 = 0 \implies x = 1.[/tex]
[tex]x + 2 = 0 \implies x = -\ 2.[/tex]
That certainly does NOT mean that plus 1 and minus 2 are the same number. What it means is that the equation has THREE different numbers that satisfy the equation , meaning to make the equation true.
[tex]0^3 + 0^2 - (2 * 0) = 0 + 0 - 0 = 0.[/tex]
[tex]1^3 + 1^2 - (2 * 1) = 1 + 1 - 2 = 0.[/tex]
[tex](-\ 2)^3 + (-\ 2)^2 - (2 * \{-\ 2\}) = -\ 8 + 4 + 4 = 0.[/tex]
There are three solutions to the equation, but those solutions are different numbers.
To render it quickly in basic English "x is zero or plus one or minus two."
An equation may have more than one solution.
Last edited by JeffM; 03-07-2018 at 04:09 PM. Reason: Minor LaTeX Error
Thank you all.
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