Integral

i have a math problem : Question :

consider the integral expression in x

P=x^3+x^2+ax+1

where a is a rational number. at a = (...........) the value of P is a rational number for any x which satisfies the equation x^2+2x-2=0 , and in this case the value of P is (.........)

please help me in answering the question on its points thak you for your help

Note by Fadlan Semeion
5 years, 2 months ago

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If \(x^2+2x-2=0\), then \(x^2=-2x+2\). Also, \((x^2+2x-2)(x-2)=x^3-6x+4=0\) so that \(x^3=6x-4\).

If you substitute these in to \(P\), you get the expression in terms of only the first degree of \(x\): \(P=x^3+x^2+ax+1=(6x-4)+(-2x+2)+ax+1=(a+4)x-1\).

You can see that by making \(a=-4\), \(P\) will be rational (specifically, \(P\) will equal \(-1\)). Note that no other rational value of \(a\) works because \((a+4)x-1\) would be a nonzero rational number (\(a+4\)) times an irrational number (\(x\)) minus a rational number (\(-1\)), which would be irrational.

Ryoma Canastra - 5 years, 2 months ago

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from what (x-2) ?

Fadlan Semeion - 5 years, 2 months ago

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Sorry that I didn't explain that. I chose to multiply \(x^2+2x-2\) by \(x-2\) because that introduces an \(x^3\) term (so that you can make the substitution) and because the result doesn't have an \(x^2\) term (which would get in the way in the substitution, although even if there were an \(x^2\) term you could use \(x^2=-2x+2\) to reduce the second degree to first degree).

Ryoma Canastra - 5 years, 2 months ago

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@Ryoma Canastra oh, i understand. i have other some math problem

this problem :

consider two conditions x^2-3x-10<0 and |x-2|<a ona real number x, where a is a positive real number.

(1) necessary and sufficient condition for x^2-3x-10<0 that (......)<x<(......)

(2) the range of values of a such that |x-2|<a is necessary condition for x^2-3x-10 is (......)

(3) the range of values of a such that |x-2|<a is sufficient condition for x^2-3x-10 is (......)

Fadlan Semeion - 5 years, 1 month ago

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