Lets suppose we have this eq: $x^2-x -2 =0$. It factors into $(x+1)(x-2)=0$, if we solve it as an second equation with Delta equals 9, we have to conclude that the equation accepts two solutions: $x_1 = -1$ and $x_2 = 2$. If we solve it as a factored polynomial, we have to conclude that the equation accepts either $ x = -1$ Or $x =2$ and these are very different conclusions if not contradictory. How should I see this?
Second degree equation solutions involving and / Or
81 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
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As everyone here is saying, but in other words maybe,
The statement "the solutions of the equation are $-1$ and $2$" is correct,
And the statement "$(x-2)(x+1)=0 \implies x=2$ or $x=-1$"is also correct. (Note that the last statement mentioned is same as saying "$x^2-x-2=0 \implies x=2$ or $x=-1$").
So the form of the equation doesn't actually matter, it just about using the words in a proper manner.
I guess (not sure if I am right) that what dazzeled you is the idea that some teachers tell students to use the word "or" when there is the equal sign $=$ and the word "and" when there is the sign $\ne$.
If so, usually teachers mean by that that if you have a fraction having $(x-2)(x+1)$ as its denominator then when asked for the domain of definition of this fraction you must say that we want $x \ne 2$ and $x \ne -1$ because both numbers give me a zero in my denominator and that is unaccepted.(i.e. we want $x$ to be different from both numbers at the same time).
So in the second statement above we can't have $x=2$ and $x=-1$ at the same time (because $x$ takes one value at a time) but here $x$ can be different than $2$ and $-1$ at the same time and that is what we want here.
Whether you solved your equation with factoring or using the quadratic formula, the equation has two solutions.
The solutions are, as you have mentioned, $x=-1$ and $x=2.$
Of course these solutions are different and there is room for confusion when we use the same notation $x$ for two different values.
In order to avoid confusion, we may use $x_1=-1$ and $x_2=2.$