Range of p satisfying quadratic inequalityWhat is the set of values of p for which p(x2+2)&lt;2x2+6x+1, for all real values of x?

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Range of p satisfying quadratic inequalityWhat is the set of values of p for which p(x2+2)&amp;lt;2x2+6x+1, for all real values of x?

Alisha Chambers · Accepted Answer

Your approach is entirely correct, and seems like the best approach to me as well. I&#039;ll write out the details here, if only for myself as I don&#039;t have any pen and paper at hand.Bringing all terms to one side, for each value of p we get a quadratic polynomial in x, and we want(2−p)x2+6x+(1−2p)&amp;gt;0,for all x. This happens if and only if 2−p&amp;gt;0 and the discriminant of the quadratic is negative, i.e.(−6)2−4(2−p)(1−2p)&amp;lt;0.The latter is a quadratic in p, and expanding the products yields−4(2p2−5p−7)&amp;lt;0.By the quadratic formula we see that this inequality holds if and only ifp&amp;lt;5−25+564=−1  or  p&amp;gt;5+25+564=72.We already found the necessary condition that 2−p&amp;gt;0, so together this shows thatp(x2+2)&amp;lt;2x2+6x+1,if and only if p&amp;lt;−1.

Range of p satisfying quadratic inequality What is the

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