How do i find a+b+c with only x1 and x2? f(x) = ax^2 + bx + c = 0, f(x1) = f(x2) = 0, x1 = 1 - sqrt[2]

AhmedAlmosuli said:

How do i find a+b+c with only x1 and x2 ?​

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First, I presume you have been told that the coefficients are all real, since that is the condition under which the two roots are conjugates.

But second, this problem has no solution unless there is some additional condition, since you can multiply all coefficients by anything, and the resulting polynomial will have the same roots. This will multiply [imath]a+b+c[/imath], changing the answer.

In particular, one polynomial with the given root is [imath]x^2-2x-1[/imath], with [imath]a+b+c=1-2-1=-2[/imath], but another is [imath]3x^2-6x-3[/imath], with [imath]a+b+c=3-6-3=-6[/imath]. These can be obtained from your results (by making different assumptions about [imath]a[/imath]), or just by writing the factored form

One more error in the problem: the first line says that [imath]f(x) = 0[/imath] as a function, which implies that [imath]a=b=c=0[/imath]. The "=0" should have been omitted, and left for the next line, where it says that [imath]x_1[/imath] and [imath]x_2[/imath] are zeros of the function.

What context have you omitted?
 
Dr.Peterson said:
First, I presume you have been told that the coefficients are all real, since that is the condition under which the two roots are conjugates.

But second, this problem has no solution unless there is some additional condition, since you can multiply all coefficients by anything, and the resulting polynomial will have the same roots. This will multiply [imath]a+b+c[/imath], changing the answer.

In particular, one polynomial with the given root is [imath]x^2-2x-1[/imath], with [imath]a+b+c=1-2-1=-2[/imath], but another is [imath]3x^2-6x-3[/imath], with [imath]a+b+c=3-6-3=-6[/imath]. These can be obtained from your results (by making different assumptions about [imath]a[/imath]), or just by writing the factored form

One more error in the problem: the first line says that [imath]f(x) = 0[/imath] as a function, which implies that [imath]a=b=c=0[/imath]. The "=0" should have been omitted, and left for the next line, where it says that [imath]x_1[/imath] and [imath]x_2[/imath] are zeros of the function.

What context have you omitted?
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Unfortunately I didn't delete any text ,the question is the same and I am with you that its wrong??
 
The other root could've also been [imath]-1-\sqrt{2}[/imath]. Not that it would change the fact the question is incomplete.
 
BigBeachBanana said:
The other root could've also been [imath]-1-\sqrt{2}[/imath]. Not that it would change the fact the question is incomplete.
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You're right; I accidentally saw it as if he were taking a complex conjugate, which is appropriate for real coefficients; taking a radical conjugate is appropriate when the coefficients are rational. Your second root would lead to an irrational coefficient, because it is not the radical conjugate; but it would still be real.

So the question is even more incomplete than I was thinking. It is common in textbooks to assume real coefficients, but not so common to assume rational (though we often give only examples with integer coefficients).

AhmedAlmosuli said:
Unfortunately I didn't delete any text ,the question is the same and I am with you that its wrong??
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I didn't say you deleted text, but that you didn't tell us the context -- if this is from a textbook, what is the chapter? That might imply some assumptions. If it is from elsewhere, where is that? In any case, I wouldn't trust anything else from that source!
 
Hello Ahmed. You've been posting faulty exercises in the forum. Do they all come from the same place?
[imath]\;[/imath]
 
For me the answer is obvious. a=b=c=0. Why? The problem clearly says that f(x)=ax^2 + bx + c =0, ie f(x) =0 (always)
 
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