You please give us at least one set of values of [imath](\alpha,\beta)[/imath] such thatthe answer of this question is B, but I don't know why.
if alpha and beta are 0, then x+1 should be 0.
so x would be -1.
but according to the question, for every x, the equation should be true.
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You please give us at least one set of values of [imath](\alpha,\beta)[/imath] such that
[imath]({\large\bf\forall}| x\in\Re)[\alpha(x+1)^2+\beta( x^2-3x - 2 )+x+1=0][/imath]
The we can discuss the matter further.
I think you missed a key part of the condition: The equation must be true for all x in R, not just for some x (-1 in your example). So your example does not contradict the claim that there are no values of alpha and beta for which the condition is true.the answer of this question is B, but I don't know why.
if alpha and beta are 0, then x+1 should be 0.
so x would be -1.
but according to the question, for every x, the equation should be true.
View attachment 28364
I know this and that is exactly why I said B cannot be the correct answer. Because if B is the correct answer, then x should be only -1.I think you missed a key part of the condition: The equation must be true for all x in R, not just for some x (-1 in your example). So your example does not contradict the claim that there are no values of alpha and beta for which the condition is true.
I hope that clarification of the problem will help you see what you need to do to solve it.
I think i get it but not sureI take it you meant
@Sarah.N:
Hint:
Can you solve your equation for [imath]\beta[/imath] in terms of [imath]\alpha[/imath] and x? If there are no solutions the x's will not cancel out. If there is one or more solutions all of the x's will cancel out. Do you see why?
-Dan
Does the question mean how many values there are for alpha and beta? I thought B says beta and alpha are zero