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

## Comments

Sort by:

TopNewestIf \(x^2+2x-2=0\), 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 · 3 years, 9 months ago

Log in to reply

(x-2)? – Fadlan Semeion · 3 years, 9 months agoLog in to reply

– Ryoma Canastra · 3 years, 9 months ago

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).Log in to reply

this problem :

consider two conditions

x^2-3x-10<0and|x-2|<aona real numberx, whereais a positive real number.(1) necessary and sufficient condition for

x^2-3x-10<0that(......)<x<(......)(2) the range of values of

asuch that|x-2|<ais necessary condition forx^2-3x-10is(......)(3) the range of values of

asuch that|x-2|<ais sufficient condition forx^2-3x-10is(......)– Fadlan Semeion · 3 years, 9 months agoLog in to reply